Rd Sharma XII Vol 1 2019 Solutions for Class 12 Commerce Math Chapter 19 Indefinite Integrals are provided here with simple step-by-step explanations. These solutions for Indefinite Integrals are extremely popular among Class 12 Commerce students for Math Indefinite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Commerce Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 19.104:

Question 1:

xx2+3x+2 dx

Answer:

     xx2+3x+2dxx=A ddxx2+3x+2+Bx=A 2x+3+Bx=2 Ax+3A+B

Comparing the Coefficients of like powers of x we get
2A=1A=123A+B=032+B=0B=-32x=12 2x+3-32

 Now, xx2+3x+2dx=122x+3-32x2+3x+2dx=122x+3dxx2+3x+2-32dxx2+3x+2=122x+3dxx2+3x+2-32dxx2+3x+322- 322+2=122x+3dxx2+3x+2 -32dxx+322-94+2=122x+3 dxx2+3x+2-32dxx+322-122=12 log x2+3x+2-32×12×12 log x+32-12x+32+12+C=12 log x2+3x+2-32 log x+1x+2+C

Page No 19.104:

Question 2:

x+1x2+x+3 dx

Answer:

x+1 dxx2+x+3x+1=Addxx2+x+3+Bx+1=A 2x+1+Bx+1 =2 Ax+A+B

Comparing Coefficients of like powers of x
2A=1A=12A+B=112+B=1B=12x+1=12 2x+1+12

 Now, x+1 dxx2+x+3=12 2x+1dxx2+x+3+12dxx2+x+3=122x+1dxx2+x+3+12dxx2+x+122- 122+3=122x+1dxx2+x+3 +12dxx+122+3 -14=122x+1 dxx2+x+3+12dxx+122+1122=12 log x2+x+3+12×211 tan-1 x+12112+C=12 log x2+x+3+111 tan-1 2x+111+C

Page No 19.104:

Question 3:

x-3x2+2x-4 dx

Answer:

x-3x2+2x-4dxx-3=Addxx2+2x-4+Bx-3=A 2x+2+Bx-3 =2 A x+2A+B

Comparing Coefficients of like powers of x

2A=1A=122A+B=-32×12+B=-3B=-4

 Now,   x-3x2+2x-4dx=122x+2 -4x2+2x-4dx=122x+2 dx x2+2x-4-4dxx2+2x+1-1-4=122x+2 dxx2+2x-4-4dxx+12- 52=12 log x2+2x-4-425 log x+1-5x+1+5+C=12 log x2+2x-4-25 log x+1-5x+1+5+C

Page No 19.104:

Question 4:

2x-3x2+6x+13 dx

Answer:

2x-3 dxx2+6x+132x-3=Addxx2+6x+13+B2x-3=A 2x+6+B2x-3 =2 A x+6A+B

Comparing Coefficients of like powers of x
2A=2A=16 A+B=-36+B=-3B=-9 2x-3=1 2x+6-9

   2x-3x2+6x+13dx=2x+6-9 x2+6x+13dx=2x+6x2+6x+13dx -9 dxx2+6x+13=2x+6 dxx2+6x+13-9dxx2+6x+32-32+13=2x+6 dxx2+6x+13-9dxx+32+22=log x2+6x+13-9×12 tan-1 x+32+C=log x2+6x+13-92 tan-1 x+32+C

Page No 19.104:

Question 5:

x-13x2-4x+3 dx

Answer:

x-13x2-4x+3dxx-1=Addx3x2-4x+3+Bx-1=A 6x-4+Bx-1 =6 A x+B-4 A

Comparing the Coefficients of like powers of x
6 A=1A=16B-4 A=-1B-4×16=-1B=-1+23B=13

 Now,  x-1 dx3x2-4x+3=166x-4+133x2-4x+3dx=166x-4 dx3x2-4x+3+13dx3x2-4x+3=166x-4 dx3x2-4x+3+19dxx2-43x+1=166x-4 dx3x2-4x+3+19dxx2-43x+232 232+1=166x-4 dx3x2-4x+3+19dxx-232-49+1=166x-4 dx3x2-4x+13+19dxx-232+532=16 log 3x2-4x+3+19×35 tan-1 x-2353+C=16 log 3x2-4x+3+135 tan-1 3 x-25+C=16 log 3x2-4x+3+515 tan-1 3x-25+C

Page No 19.104:

Question 6:

2x2+x-x2 dx

Answer:

2x dx2+x-x22x=Addx2 +x-x2+B2x=A 0+1-2x+B2x=-2 A x+A+B

Comparing the Coefficients of like powers of x
-2 A=2A=-1A+B=0-1+B=0B=1

 Now,   2x dx2+x-x2=-11-2x+1-x2+x+2dx=-1-2x-x2+x+2dx+dx-x2+x+2=-I1+I2         ...   1     say whereI1=1-2x-x2+x+2dxI2=dx-x2+x+2I1=1-2x-x2+x+2dxlet -x2+x+2=t1-2x dx=dtI1=dttI1=log t+C1=log 2+x-x2+C1          ...   2I2=dx-x2+x+2I2=-dxx2-x-2I2=-dxx2-x+122-122-2I2=-dxx-122-322I2=-12×32log x-12-32x-12+32+C2I2=-13 log x-2x+1+C2       ...    3from 1 2 and 32x2+x-x2dx=-log 2+x-x2-13log x-2x+1+C1+C2=-log 2+x-x2+13 log 1+xx-2+Cwhere C =C1+C2

Page No 19.104:

Question 7:

1-3x3x2+4x+2 dx

Answer:

1-3x dx3x2+4x+21-3x=Addx3x2+4x+2+B1-3x=A 6x+4+B1-3x=6 A x+4 A+B

Comparing the Coefficients of like powers of x

6 A=-3A=-124 A+B=14×-12+B=1B=3

 1-3x=-126x+4+3Now,   1-3x dx3x2+4x+2=-126x+4+33x2+4x+2dx=-126x+4 dx3x2+4x+2+3dx3x2+4x+2=-12 I1+3I2         say    ...   1whereI1=6x+43x2+4x+2 and I2=dx3x2+4x+2I1=6x+43x2+4x+2dxlet 3x2+4x+2=t6x+4 dx=dtI1=dtt=log t+C1=log 3x2+4x+2+C1        ...       2I2=dx3x2+4x+2I2=13dxx2+43x+23I2=13dxx2+4xx+232-232+23I2=13dxx-232-49+23I2=13dxx+232+29I2=13dxx+232+232I2=13×32 tan-1 x+2323+C2I2=12 tan-1 3x+22+C2          ...      3from 1, 2 and 31-3x dx3x2+4x+2=-12 log 3x2+4x+2+3×12 tan-1 3x+22+C1+C2=-12 log 3x2+4x+2+32 tan-1 3x+22+C        Where C=C1+C2

Page No 19.104:

Question 8:

2x+5x2-x-2 dx

Answer:

2x+5 dxx2-x-22x+5=Addxx2-x-2+B2x+5=A 2x-1+B2x+5=2 A x+B-A

Comparing the Coefficients of like powers of x

2 A=2A=1B-A=5B-1=5B=6

  2x+5=1·2x-1+6  2x+5x2-x-2dx2x-1+6x2-x-2dx2x-1x2-x-2dx+6dxx2-x-2=I1+6 I2         say    ...   1whereI1=2x-1x2-x-2dx I2=dxx2-x-2I1=2x-1x2-x-2dxlet x2-x-2=t2x-1 dx=dtI1=dttI1=log tI1=log x2-x-2+C1      ...       2I2=dxx2-x-2I2=dxx2-x+122- 122-2I2=dxx-122-14-2I2=dxx-122-322I2=12×32 log x-12-32x-12+32I2=13 log x-2x+1+C2         ...      32x+5 dxx2-x-2=log x2-x-2+63 log x-2x+1+C1+C2=log x2-x-2+2 log x-2x+1+C     Where C=C1+C2  

Page No 19.104:

Question 9:

ax3+bxx4+c2 dx

Answer:

    ax3+bx x4+c2dx=ax3 x4+c2dx+bx x22+c2dx=I1+I2     sayWhereI1=ax3 x4+c2dx    &   I2=bx x22+c2dxNow, I1=ax3 x4+c2dxlet x4+c2=t4x3 dx=dtx3 dx=dt4I1=a4dtt=a4 log t+C1=a4 log x4+c2+C1Now, I2=bx x22+c2dxlet x2=p2x dx=dpx dx=dp2

 I2=b2dpp2+c2=b2×1c tan-1 pc+C2=b2c tan-1 x2c+C2ax3+bx x4+c2dx=a4 log x4+c2+b2c tan-1 x2c+C1+C2                        =a4 log x4+c2+b2c tan-1 x2c+C    Where C=C1+C2

Page No 19.104:

Question 10:

3 sin x-2 cos x5-cos2 x-4 sin x dx

Answer:

    3 sin x-2 cos x dx5-cos2x-4 sin x=3 sin x-2 cos x dx5-1-sin2x-4 sin x=3 sin x-2 cos x dx sin2x-4 sin x+4Let sin x=tcos x dx=dt3t-2 dtt2-4t+43t-2=Addxt2-4t+4+B3t-2=A 2t-4+B3t-2=2 A t+B-4 A

Comparing the Coefficients of like powers of t

2 A=3A=32B-4 A=-2B-4×32=-2B=-2+6B=4

  3t-2=32 2t-4+4 3t-2 dtt2-4t+4=322t-4+4t2-4t+4dt=322t-4t2-4t+4dt+4dtt2-4t+4=32 I1+4 I2       ...   1whereI1=2t-4 dtt2-4t+4, I2=dtt2-4t+4I1=2t-4 dtt2-4t+4Let t2-4t+4=p2t-4 dt=dpI1=2t-4 dtt2-4t+4=dpp=log p+C1=log t2-4t+4+C1        ...       2I2=dtt2-4t+4I2=dtt-22I2=t-2-2 dtI2=t-2-2+1-2+1+C2I2=-1t-2+C2          ...      3from 1, 2 and 33 sin x-2 cos x dx5-cos2x-4 sinx=32 log t2-4t+4+4×-1t-2+C1+C2=32 log sin2x-4 sin x+4+42-t+C                      Where C=C1+C2=32log sin x-22+42-sin x+C=32×2 log sin x-2+42-sin x+C=3 log 2-sin x+42-sin x+C

Page No 19.104:

Question 11:

x+22x2+6x+5 dx

Answer:

x+22x2+6x+5dxx+2=Addx2x2+6x+5+Bx+2=A 4x+6+Bx+2=4 A x+6 A+B

Comparing the Coefficients of like powers of x

4 A=1A=146 A+B=26×14+B=2B=12

   x+22x2+6x+5dx=144x+6+122x2+6x+5dx=144x+62x2+6x+5dx+1212x2+6x+5dx=144x+62x2+6x+5dx+14dxx2+3x+52=144x+62x2+6x+5dx+14dxx2+3x+322-322+52=144x+62x2+6x+5dx+14dxx+322-94+52=144x+62x2+6x+5dx+14dxx+322+14=144x+62x2+6x+5dx+14dxx+322+122=14 log 2x2+6x+5+14×2 tan-1 x+3212+C=14 log 2x2+6x+5+12 tan-1 2x+3+C

Page No 19.104:

Question 12:

Evaluate the following integrals:

5x-21+2x+3x2dx

Answer:

Let I=5x-21+2x+3x2dx       =5x-23x2+2x+1dxWe express 5x-2=Addx3x2+2x+1+B5x-2=A(6x+2)+BEquating the coefficients of x and constants, we get5=6A      and     -2=2A+Bor A=56      and     B=-113 I=566x+2-1133x2+2x+1dx     =566x+23x2+2x+1dx-11313x2+2x+1dx     =56I1-113I2       ...(1)Now, I1=6x+23x2+2x+1dx      Let 3x2+2x+1=t      On differentiating both sides, we get      6x+2dx=dt I1=1tdt       =logt+c1       =log3x2+2x+1+c1      ...(2)And, I2=13x2+2x+1dx          =131x2+23x+13dx          =131x2+23x+19-19+13dx          =131x+132+232dx      Let x+13=t      On differentiating both sides, we get      dx=dt I2=131t2+232dt       =13×123tan-13t2+c2       =12tan-13x+132+c2       =12tan-13x+12+c2      ...(3)From (1), (2) and (3), we get I=56log3x2+2x+1+c1-11312tan-13x+12+c2      =56log3x2+2x+1-11312tan-13x+12+cHence, 5x-21+2x+3x2dx=56log3x2+2x+1-11312tan-13x+12+c

Page No 19.104:

Question 13:

x+53x2+13x-10dx

Answer:

I=x+53x2+13x-10dx=x+53x2+15x-2x-10dx=x+53xx+5-2x+5dx=x+53x-2x+5dx

=x+5(3x-2)(x+5)dx=13x-2dx I=13ln3x-2+c

Page No 19.104:

Question 14:

3sin x-2cos x13-cos2 x-7sin xdx

Answer:

I=3sin x-2cos x13-cos2 x-7sin xdx

=3sin x-2 cos x13-(1-sin2x)-7sinx dx             cos2x=1-sin2x=3sin x-2 cos xsin2 x-7sin x+12dx=3sin x-2 cos xsin2 x-4sin x-3sin x+12dx=3sin x-2 cos xsin xsin x-4-3sin x-4dx= 3sin x-2cos xsin x-3sin x-4dx

Let sin x=tcos x dx=dt I=3t-2t-3t-4dt

Using partial fraction, we get

3t-2t-3t-4=At-3+Bt-4=At-4+Bt-3t-3t-43t-2=(A+B)t-4A-3B

Comparing coefficients, we get

A = -7 and B = 10

So, I=-71t-3dt+101t-4dt

I=-7lnt-3+10lnt-4 +c I=-7lnsin x-3+10 lnsin x-4+c

Page No 19.104:

Question 15:

x+73x2+25x+28dx

Answer:

I=x+73x2+25x+28dx=x+73x2+21x+4x+28dx=x+73xx+7+4x+7dx=x+73x+4x+7dx

=1(3x+4)dx=13ln3x+4+c

Page No 19.104:

Question 16:

Evaluate the following integrals:
x3x4+x2+1dx

Answer:

I=x3x4+x2+1dx=x2·xx22+x2+1dxLet x2=t or 2xdx=dtI=12tt2+t+1dt=142tt2+t+1dt=142t+1-1t2+t+1dt

=142t+1t2+t+1-1t2+t+1dt=14logt2+t+1-1t2+t+14+34dt=14logt2+t+1-1t+122+322dt=14logt2+t+1-23tant+1232+c=14logt2+t+1-23tan2t+13+c

=14logx4+x2+1-23tan2x2+13+c

Page No 19.104:

Question 17:

Evaluate the following integrals:
x3-3xx4+2x2-4dx

Answer:

I=x3-3xx4+2x2-4dx

=x(x2 -3)x4+2x2-4dx

Let x2=t, or, 2xdx=dt

I=12(t-3)t2+2t-4dt=142t-6t2+2t-4dt=142t+2-8t2+2t-4dt=142t+2t2+2t-4-8t2+2t-4dt=142t+2t2+2t-4dt-8t2+2t-4dt

I=14I1+I2        ...i

Now,

I1  = 2t + 2t2 + 2t -4 dt

Let t2+2t-4=u

or, 2t+2dt=duI1=1u du=lnu+c1I1=lnt2+2t-4+c1 I1=lnx4+2x2-4+ c1

Now,

I2=-8(t+1)2-5dtI2=8(5)2-(t+1)2dt I2=825ln5+x2+15-x2-1+c2

So,from i, we getI=14lnx4+2x2-4+45ln 5+x2+15-x2-1+C I=14lnx4+2x2-4+15ln 5+x2+15-x2-1+C



Page No 19.106:

Question 1:

x2+x+1x2-x dx

Answer:

x2+x+1x2-xdxx2+x+1x2-x=1+2x+1x2-xx2+x+1x2-xdx=1+2x+1x2-xdx=1+2x-1+2x2-xdx=dx+2x-1 dxx2-x+2 dxx2-x+122-122=dx+2x-1 dxx2-x+2dxx-122-122=x+log x2-x+2×12×12logx-12-12x-12+12=x+log x2-x+2 log x-1x+C

Page No 19.106:

Question 2:

x2+x-1x2+x-6 dx

Answer:

x2+x-1x2+x-6dxx2+x-1x2+x-6=1+5x2+x-6    x2+x-1x2+x-6dx=dx+5dxx2+x-6=dx+5dxx2+x+122-122-6=dx+5dxx+122-14-6=dx+5dxx+122-522=x+5×12×52 log x+12-52x+12+52+C=x+log x-2x+3+C

Page No 19.106:

Question 3:

1-x2x 1-2x dx

Answer:

We have,I=1-x2 x 1-2xdx=-x2+1-2x2+xdx=12dx+1-x2-2x2+xdx=12dx+122-x-2x2+xdx=12dx+2-x-2x2+xdx=12I1+I2          saywhere I1=dx    &   I2=2-x-2x2+xdxNow, I1=dx             =x+C1I2=2-x-2x2+xdxLet 2-x=A ddx -2x2+x+B2-x=A -4x+1+B2-x=-4Ax+A+B

Comparing coefficients of like terms

-1=-4 A A=14& A+B=214+B=2B=2-14       =8-14       =74

2-x-2x2+xdx=14-4x+1+74-2x2+xdx                           =14-4x+1-2x2+xdx+74-2x2+xdx                           =14log -2x2+x+741-2x2-x2+116-116dx                           =14log -2x2+x-781x-142-142dx                           =14log x-2x+1-78×12×14logx-14-14x-14+14+C2                           =14log x+14log -2x+1-74logx-12x+C2                           =14log x+14log -2x+1-74log x-12+74log x+C2                           =2 log x+14log -2x+1-74log 2x-1+C3  , where C3=C2+74log 2                           =2 log x+14log -2x+1-74log 1-2x+C3                           =2 log x-32log 1-2x+C3Thus, I=  12x+C1+2 log x-32log 1-2x+C3          =12x+log x-34log 1-2x+C,        where C=12C1+C3

Page No 19.106:

Question 4:

x2+1x2-5x+6 dx

Answer:

Let Ix2+1x2-5x+6dxDividing Numerator by Denominatorx2-5x+6x2+1     1                   x2-5x+6                 -  +      -                       5x-5x2+1x2-5x+6=1+5x-5x2-5x+6          ..... 1Also 5x-5x2-5x+6=5x-5x-2  x-3Let 5x-5x-2  x-3=Ax-2+Bx-35x-5x-2 x-3=A x-3 +B x-2x-2  x-35x-5=A x-3+B x-2let x=35×3-5=A×0+B 3-210=Blet x=25×2-5=A 2-3+B×0A=-55x-5x-2 x-3=-5x-2+10x-3         .....2from 1 and 2I=dx-5dxx-2+10dxx-3=x-5 log x-2+10 log x-3+C

Page No 19.106:

Question 5:

x2x2+7x+10 dx

Answer:

Let I=x2x2+7x+10dxNow,x2+7x+10x2            1                     x2+7x+10                   -  -      -                        -7x-10 x2x2+7x+10=1-7x+10x2+7x+10x2x2+7x+10=1-7x+10x2+2x+5x+10x2x2+7x+10=1-7x+10x x+2 +5 x+2x2x2+7x+10=1- 7x+10x+2 x+5         ..... 1Consider,7x+10x+2  x+5=Ax+2+Bx+57x+10=A x+5+B x+2let x+5=0x=-57 -5+10=A×0+B -5+2-25=B -3B=253let x+2=0x=-27 -2+10=A -2+5-4=A 3A=-437x+10x+2 x+5=-43 x+2+253 x+5         .....2from 1 and 2x2x2+7x+10=1+43 x+2-253 x+5x2 dxx2+7x+10=dx+43dxx+2-253dxx+5=x+43 log x+2-253 log x+5+C

Page No 19.106:

Question 6:

x2+x+1x2-x+1 dx

Answer:

Let I=x2+x+1x2-x+1dxNow,x2-x+1x2+x+1  1                  x2-x+1                 -  +  -                             2x          Therefore,x2+x+1x2-x+1=1+2xx2-x+1x2+x+1x2-x+1 dx=dx+2x-1+1x2-x+1 dx=dx+2x-1x2-x+1 dx+dxx2-x+1=dx+2x-1 dxx2-x+1+dxx2-x+122-122+1=dx+2x-1 dxx2-x+1+dxx-122+322=x+ log x2-x+1+23 tan-1 2x-13+C

Page No 19.106:

Question 7:

x-12x2+2x+2 dx

Answer:

Let I=x-12x2+2x+2 dx=x2-2x+1x2+2x+2 dxHere,x2+2x+2x2-2x+1  1                  x2+2x+2                  -  -      -                   -4x-1   Therefore,x2-2x+1x2+2x+2=1-4x+1x2+2x+2        ..... 1Let 4x+1=Addx x2+2x+2+B4x+1=A 2x+2+B4x+1=2A x+2A+BEquating Coefficients of like terms2A=4A=22A+B=12×2+B=1B=-3x2-2x+1x2+2x+2 dx=dx-22x+2x2+2x+2 dx+3dxx2+2x+2=dx-22x+2x2+2x+2 dx+3dxx+12+12=x-2 log x2+2x+2+31 tan-1 x+11+C=x-2 log x2+2x+2+3 tan-1 x+1+C

Page No 19.106:

Question 8:

x3+x2+2x+1x2-x+1 dx

Answer:

Let I=x3+x2+2x+1x2-x+1 dxx2-x+1x3+x2+2x+1  x+2                  x3-x2+x                 -  +   -                             2x2+ x+1                     2x2-2x+2                     -  +    -                               3x-1Therefore,x3+x2+2x+1x2-x+1=x+2+3x-1x2-x+1        ..... 1Let3x-1=Addx x2-x+1+B3x-1=A 2x-1+B3x-1=2A x+B-AEquating Coefficients of like terms2A=3A=32B-A=-1B-32=-1B=12x3+x2+2x+1x2-x+1 dx =x+2 dx+32 2x-1+12x2-x+1 dx=x+2 dx+32 2x-1x2-x+1 dx+12dxx2-x+1=x+2 dx+322x-1 dxx2-x+1+12dxx2-x+14-14+1=x+2 dx+322x-1 dxx2-x+1 +12dxx-122+322=x22+2x+32 logx2-x+1+12×23 tan-1 x-1232+C=x22+2x+32 log x2-x+1+13 tan-1 2x-13+C

Page No 19.106:

Question 9:

x2 x4+4x2+4 dx

Answer:

  Let I=x2 x4+4x2+4 dx=x6+4x2x2+4 dxNow,x2+4x6+4x2  x4-4x2+20           x6+4x4                            -   -                           -4x4+4x2            -4x4-16x2            +     +                          20x2                20x2 + 80               -        -                       -80           Therefore, x2 x4+4x2+4=x4-4x2+20-80x2+4I=x2 x4+4x2+4 dx=x4-4x2+20 dx-80dxx2+ 22=x4 dx-4x2 dx+20dx-80dxx2+22=x4+14+1-4 x33+20 x-80×12 tan-1 x2+C=x55-43 x3+20x-40 tan-1 x2+C

Page No 19.106:

Question 10:

x2x2+6x+12 dx

Answer:

Let I=x2 dxx2+6x+12Now,x2+6x+12x2                1                      x2+6x+12                      -  -    -                               -6x-12Therefore,x2x2+6x+12=1-6x+12x2+6x+12        ..... 1Let 6x+12=Addx x2+6x+12+B6x+12=A 2x+6+B6x+12=2A x+6A+BEquating Coefficients of like terms2A=6A=36A+B=1218+B=12B=-6x2x2+6x+12=1-3 2x+6-6x2+6x+12I=x2 dxx2+6x+12=dx-32x+6 dxx2+6x+12+6dxx2+6x+12=dx-3 2x+6 dxx2+6x+12+6dxx2+6x+9+3=dx-32x+6 dxx2+6x+12+6dxx+32+32=x-3 log x2+6x+12+63 tan-1 x+33+C=x-3 log x2+6x+12+23 tan-1 x+33+C



Page No 19.110:

Question 1:

xx2+6x+10 dx

Answer:

Let I=x dxx2+6x+10x=A ddx x2+6x+10+Bx=A 2x+6+Bx=2A x+6A+BEquating Coefficients of like terms2A=1A=126A+B=06×12+B=0B=-3I=x dxx2+6x+10=12 2x+6-3x2+6x+10dx=122x+6 dxx2+6x+10-3dxx2+6x+32-32+10=122x+6 dxx2+6x+10-3dxx+32+12let x2+6x+10=t2x+6 dx=dtI=12dtt-3dxx+32+1=12×2t-3 log x+3+x+32+1+C=t-3 log x+3+x2+6x+10+C=x2+6x+10-3 log x+3+x2+6x+10+C

Page No 19.110:

Question 2:

2x+1x2+2x-1 dx

Answer:

Let I=2x+1 dxx2+2x-1=2x+2-1 dxx2+2x-1=2x+2 dxx2+2x-1-dxx2+2x-1=2x+2 dxx2+2x-1-dxx2+2x+1-1-1=2x+2 dxx2+2x-1-dxx+12-22let x2+2x-1=t2x+2 dx=dtI=dtt-dxx+12-22=2t-log x+1+x+12-22+C=2x2+2x-1-log x+1+x2+2x-1+C

Page No 19.110:

Question 3:

x+14+5x-x2 dx

Answer:

Let I=x+1 dx4+5x-x2Also, x+1=A ddx 4+5x-x2+Bx+1=A 5-2x+Bx+1=-2A x+5A+BEquating Coefficients of like terms-2A=1A=-12And5A+B=1-52+B=1B=72I=x+1 dx4+5x-x2=-12 5-2x+724+5x-x2dx=-125-2x dx4+5x-x2+72dx4-x2-5x=-125-2x dx4+5x-x2+72dx4-x2-5x+522-522=-125-2x dx4+5x-x2+72dx4-x-522+254=-125-2x4+5x-x2dx+72dx414-x-522=-125-2x4+5x-x2dx+72dx4122-x-522let 4+5x-x2=t5-2x dx=dtThen,I=-12dtt+72dx4122-x-522=-12×2t+72×sin-1 x-52412+C=-t+72 sin-1 2x-541+C=-4+5x-x2+72 sin-1 2x-541+C

Page No 19.110:

Question 4:

6x-53x2-5x+1 dx

Answer:

Let I=6x-53x2-5x+1dxPutting 3x2-5x+1=t6x-5 dx=dtThen,I=dtt=2t+C=23x2-5x+1+C

Page No 19.110:

Question 5:

3x+15-2x-x2 dx

Answer:

Let I=3x+1 dx5-2x-x2Consider, 3x+1=A ddx 5-2x-x2+B3x+1=A -2-2x+B3x+1=-2A x-2A+BEquating Coefficients of like terms-2A=3A=-32And-2A+B=1-2×-32+B=1B=1-3B=-2I=-32 -2-2x-25-2x-x2dx=-32-2-2x dx5-2x-x2-2dx5-2x-x2=-32-2-2x dx5-2x-x2-2dx5-x2+2x=-32-2-2x dx5-2x-x2-2dx5-x2+2x+1-1=-32-2-2x dx5-2x-x2-2dx6-x+12=-32-2-2x dx5-2x-x2-2dx62-x+12let 5-2x-x2=t-2-2x dx=dtI=-32dtt-2dx62-x+12=-32×2t-2 sin-1 x+16+C=-35-2x-x2-2 sin-1 x+16+C

Page No 19.110:

Question 6:

x8+x-x2 dx

Answer:

Let I=x dx8+x-x2Consider, x=Addx 8+x-x2+Bx=A 1-2x+Bx=-2A x+A+BEquating Coefficients of like terms-2A=1A=-12AndA+B=0-12+B=0B=12x=-12 1-2x+12Then,I=-121-2x dx8+x-x2+12dx8+x-x2=-121-2x dx8+x-x2+12dx8-x2-x=-121-2x dx8+x-x2+12dx8-x2-x+14-14=-121-2x dx8+x-x2+12dx8+14-x-122=-121-2x dx8+x-x2+12dx3322-x-122let 8+x-x2=t1-2x dx=dtI=-12dtt+12dx3322-x-122=-12×2t+12 sin-1 x-12332+C=-t+12 sin-1 2x-133+C=-8+x-x2+12 sin-1 2x-133+C

Page No 19.110:

Question 7:

x+2x2+2x-1 dx

Answer:

​Let I=x+2 dxx2+2x-1Consider,x+2=A ddx x2+2x-1+Bx+2=A 2x+2+Bx+2=2A x+2A+BEquating Coefficients of like terms2A=1A=12And2A+B=22×12+B=2B=1Then,I=12 2x+2+1x2+2x-1dx=122x+2 dxx2+2x-1+dxx2+2x-1let x2+2x-1=t2x+2 dx=dtI=12dtt+dxx2+2x-1=12t-12dt+dxx2+2x+1-2=12 t-12+1-12+1+dxx+12-22=t+ log x+1+x+12-22+C=x2+2x-1+ log x+1+x2+2x-1+C

Page No 19.110:

Question 8:

x+2x2-1 dx

Answer:

​Let I=x+2x2-1dx=xx2-1dx+2dxx2-1let x2-1=t2x dx=dtx dx=dt2Then,I=12dtt+2dxx2-12=12t-12 dt+2dxx2-12=12 t-12+1-12+1+2 log x+x2-1+C=t+2 log x+x2-1+C=x2-1+2 log x+x2-1+C

Page No 19.110:

Question 9:

x-1x2+1 dx

Answer:

​Let I=x-1x2+1 dx=x dxx2+1-dxx2+1Putting x2+1=t2x dx=dtx dx=dt2Then,I=12dtt-dxx2+12=12t-12dt-dxx2+12=12 t-12+1-12+1-dxx2+12=t- log x+x2+1+C=x2+1- log x+x2+1+C

Page No 19.110:

Question 10:

xx2+x+1 dx

Answer:

Let I=x dxx2+x+1Consider,x=A ddx x2+x+1+Bx=A 2x+1+Bx=2A x+A+BEquating Coefficient of like terms2A=1A=12AndA+B=012+B=0B=-12I=12 2x+1-12x2+x+1 dx=122x+1x2+x+1dx-12dxx2+x+14-14+1Putting x2+x+1=t2x+1 dx=dtThen,I=12dtt-12dxx+122+322=12t-12 dt-12 log x+12+x+122+322+C=12t-12+1-12+1-12 log x+12+x2+x+1+C=t-12 log x+12+x2+x+1+C=x2+x+1-12 log x+12+x2+x+1+C

Page No 19.110:

Question 11:

x+1x2+1 dx

Answer:

​Let I=x+1x2+1 dx=x dxx2+1+dxx2+1Putting, x2+1=t2x dx=dtx dx=dt2Then,I=12dtt+dxx2+1=12t-12dt+dxx2+1=12 t-12+1-12+1+log x+x2+1+C=t+ log x+x2+1+C=x2+1+ log x+x2+1+C

Page No 19.110:

Question 12:

2x+5x2+2x+5 dx

Answer:

Let I=2x+5 dxx2+2x+5Consider,2x+5=A ddx x2+2x+5+B2x+5=A 2x+2+B2x+5=2A x+2A+BEquating Coefficients of like terms2A=2A=1And 2A+B=5B=3I=2x+2+3x2+2x+5 dx=2x+2 dxx2+2x+5+3dxx2+2x+5let x2+2x+5=t2x+2 dx=dtThen,I=dtt+3dxx2+2x+1+4=t-12 dt+3 dxx+12+22=t-12+1-12+1+3 log x+1+x+12+4+C=2t+3 log x+1+x2+2x+5+C=2x2+2x+5+3 log x+1+x2+2x+5+C

Page No 19.110:

Question 13:

3x+15-2x-x2 dx

Answer:

Let I=3x+1 dx5-2x-x2Consider,3x+1=A ddx 5-2x-x2+B3x+1=A -2-2x+B3x+1=-2A x+-2A+BEquating Coefficients of like terms-2A=3A=-32And-2A+B=1-2×-32+B=1B=-2I=-32 -2-2x-25-2x-x2 dx=-32-2-2x dx5-2x-x2-2dx5-2x-x2=-32-2-2x dx5-2x-x2-2dx5-x2+2x=-32-2-2x dx5-2x-x2-2dx5-x2+2x+1-1=-32-2-2x dx5-2x-x2-2 dx6-x+12=-32-2-2x dx5-2x-x2-2dx62-x+12Putting, 5-2x-x2=t-2-2x dx=dtThen,I=-32dtt-2 sin-1 x+16+C1=-32×2t-2 sin-1 x+16+C=-35-2x-x2-2 sin-1 x+16+C

Page No 19.110:

Question 14:

1-x1+x dx

Answer:

Let I=1-x1+x dx=1-x1-x1+x1-x dx=1-x1-x2 dx=dx1-x2-x dx1-x2Putting 1-x2=t-2x dx=dtx dx =-dt2Then,I=dx1-x2+12dtt=sin-1 x+12×2t+C=sin-1 x+1-x2+C



Page No 19.111:

Question 15:

2x+1x2+4x+3 dx

Answer:

Let I=2x+1 dxx2+4x+3Consider,2x+1=A ddx x2+4x+3+B2x+1=A 2x+4+B2x+1=2A x+4A+BEquating Coefficients of like terms2A=2A=1And4A+B=14+B=1B=-3I=2x+4-3x2+4x+3dx=2x+4 dxx2+4x+3 -3dxx2+4x+4-4+3=2x+4 dxx2+4x+3-3dxx+22-12Let x2+4x+3=t2x+4 dx=dtThen,I=dtt-3dxx+22-12=t-12 dt-3 dxx+22-12=t-12+1-12+1-3 log x+2+x+22-1+C=2t-3 log x+2+x2+4x+3+C=2x2+4x+3-3 log x+2+x2+4x+3+C

Page No 19.111:

Question 16:

2x+3x2+4x+5 dx

Answer:

​Let I=2x+3 dxx2+4x+5=2x+4-1x2+4x+5dx=2x+4 dxx2+4x+5-dxx2+4x+5=2x+4 dxx2+4x+5-dxx+22+1Consider, x2+4x+5=t2x+4 dx=dtI=dtt-dxx+22+12=t-12 dt-dxx+22+12=t-12+1-12+1- log x+2+x+22+1+C=2x2+4x+5- log x+2+x2+4x+5+C

Page No 19.111:

Question 17:

5x+3x2+4x+10 dx

Answer:

Let I=5x+3 dxx2+4x+10Consider,5x+3=A ddx x2+4x+10+B5x+3=A 2x+4+B5x+3=2A x+4A+BEquating Coefficients of like terms2A=5A=52And4A+B=34×52+B=3B=-7I=522x+4 dxx2+4x+10-7dxx2+4x+10=522x+4 dxx2+4x+10 -7dxx2+4x+4-4+10=522x+4 dxx2+4x+10-7dxx+22+62Putting, x2+4x+10=t2x+4 dx=dtThen,I=52dtt-7 log x+2+x+22+6+C=52t-12 dt-7 log x+2+x2+4x+10+C=52×2t-7 log x+2+x2+4x+10+C=5x2+4x+10-7 log x+2+x2+4x+10+C

Page No 19.111:

Question 18:

Evaluate the following integrals:

x+2x2+2x+3dx

Answer:

Let I=x+2x2+2x+3dxWe express x+2=Addxx2+2x+3+Bx+2=A(2x+2)+BEquating the coefficients of x and constants, we get1=2A      and     2=2A+Bor A=12      and     B=1I=122x+2+1x2+2x+3dx     =122x+2x2+2x+3dx+1x2+2x+3dx     =12I1+I2       ...(1)Now, I1=2x+2x2+2x+3dx      Let x2+2x+3=u      On differentiating both sides, we get      2x+2dx=du I1=1udu       =2u+c1       =2x2+2x+3+c1      ...(2)And, I2=1x2+2x+3dx          =1x2+2x+1-1+3dx          =1x+12+22dx      Let x+1=u      On differentiating both sides, we get      dx=du I2=1u2+22du       =logu+u2+22+c2       =logx+1+x2+2x+3+c2       ...(3)From (1), (2) and (3), we get I=122x2+2x+3+c1+logx+1+x2+2x+3+c2      =x2+2x+3+logx+1+x2+2x+3+cHence, x+2x2+2x+3dx=x2+2x+3+logx+1+x2+2x+3+c



Page No 19.114:

Question 1:

14 cos2 x+9 sin2 x dx

Answer:

Let I = 14 cos2 x+9 sin2 xdxDividing numerator and denominator by cos2 xI= 1cos2 x4+9 tan2 xdx      = sec2 x 4+9 tan2 xdxLet tan x=tsec2 x dx=dtI = dt4+9t2      =19 dt49+t2      =19 dt232+t2      =19×32tan-1 t23+C      =16tan-1 3t2+C      =16tan-1 3 tan x2+C

Page No 19.114:

Question 2:

14 sin2 x+5 cos2 x dx

Answer:

Let I=14 sin2 x+5 cos2 xdxDividing numerator & denominator by cos2 xI=sec2 x 4 tan2 x+5dxLet tan x=tsec2 x dx=dtI= dt4t2+5      =14 dtt2+54      =14dtt2+522      =14×25 tan-1 t5×2+C      =125tan-1 2 tan x5+C

Page No 19.114:

Question 3:

22+sin 2x dx

Answer:

Let I= 2 2+sin 2xdx        = 2 2+2 sin x cos xdx        = 11+sin x cos xdxDividing numerator and denominator by cos2 xI= sec2 x dxsec2 x+tan x      = sec2 x dx1+tan2 x+tan xLet tan x=tsec2 x dx=dtI= dtt2+t+1      =dtt2+t+14-14+1      = dtt+122+322      =23tan-1 t+1232+C      =23tan-1 2t+13+C      =23tan-1 2 tan x+13+C

Page No 19.114:

Question 4:

cos xcos 3x dx

Answer:

Let I= cos x cos 3xdx       =cos x4 cos3x-3 cos xdx            cos 3A=4 cos3 A-3 cos A       = 14 cos2 x-3dxDividing numerator and denominator by cos2 xI= sec2 x4-3 sec2 x dx      = sec2 x4-31+tan2 x dx      = sec2 x1-3 tan2 x dx      = sec2 x 1-3 tan x2 dxLet 3 tan x=t3 sec2 x dx=dtsec2 x dx=dt3I=13  dt12-t2      =13×12ln 1+t1-t+C      =123ln 1+3 tan x1-3 tan x+C

Page No 19.114:

Question 5:

11+3 sin2 x dx

Answer:

Let I= 11+ 3 sin2 xdxDividing numerator and denominator by cos2 xI= sec2 xsec2 x+3 tan2 xdx      = sec2 x 1+tan2 x+3 tan2 xdx      = sec2 x 1+4 tan2 xdx      = sec2 x 1+2 tan x2dxLet 2 tan x=t2 sec2 x dx=dtsec2 x dx=dt2I=12 dt1+t2      =12 tan-1 t+C      =12 tan-1 2 tan x+C

Page No 19.114:

Question 6:

13+2 cos2 x dx

Answer:

Let I= 13+2 cos2 xdxDividing numerator and denominator by cos2 xI= sec2 x3 sec2 x+2 dx      = sec2 x 3 1+tan2 x+2dx      = sec2 x 3 tan2 x+5dx      = sec2 x 52+3 tan x2dxLet 3 tan x=t3 sec2 x dx=dtsec2 x dx=dt3I=13 dt52+t2      =13×15 tan-1 t5+C      =115 tan-1 3 tan x5+C

Page No 19.114:

Question 7:

1sin x-2 cos x2 sin x+cos x dx

Answer:

Let I= 1sin x-2 cos x 2 sin x+cos xdxDividing numerator and denominator by cos2 xI= sec2 x sin x-2 cos xcos x×2 sin x+cos xcos xdx      = sec2 x tan x-2 2 tan x+1dxLet tan x=tsec2 x dx=dtI= dtt-2 2t+1      = dt2t2+t-4t-2      = dt2t2-3t-2      =12 dtt2-32t-1      =12 dtt2-32t+342-342-1      =12 dtt-342-916-1      =12 dtt-342-542      =12×12×54 log t-34-54t-34+54+C      =15 ln t-2t+12+C      =15ln t-222t+1+C      =15ln t-22t+1+15 ln 2+C      =15 ln t-22t+1+C'     where C'=C+15ln 2      =15 ln tan x-22 tan x+1+C

Page No 19.114:

Question 8:

sin 2xsin4 x+cos4 x dx

Answer:

Let I= sin 2xsin4 x+cos4 x dx      = 2 sin x cos xsin4 x+cos4 x dxDividing numerator & denominator by cos4 xI= 2 sin x cos xcos4 xtan4 x+1dx      = 2 tan x. sec2 xtan2 x2+1 dxLet tan2 x=t2 tan x sec2 x dx=dt= dtt2+1I=tan-1 t+C      =tan-1 tan2 x+C

Page No 19.114:

Question 9:

1cos x sin x+2 cos x dx

Answer:

Let I= 1cos xsin x+2 cos xdxDividing numerator and denominator by cos2 xI= sec2 x cos xcos x×sin x+2 cos xcos xdx      = sec2 x tan x+2dxLet tan x+2=tsec2 x dx=dtI= dtt      =ln t+C      =ln tan x+2+C

Page No 19.114:

Question 10:

1sin2 x+sin 2x dx

Answer:

Let I= 1sin2 x+sin 2xdx      = 1sin2 x+2 sin x cos xdxDividing numerator and denominator by cos2 xI= sec2 x tan2 x+2 tan xdxLet tan x=tsec2 x dx=dt I= dtt2+2t      = dtt2+2t+1-1      = dtt+12--12      =12ln t+1-1t+1+1+C      =12ln tt+2+C      =12ln tan xtan x+2+C

Page No 19.114:

Question 11:

1cos 2x+3 sin2 x dx

Answer:

Let I= 1cos 2x+3 sin2 xdx      = 11-2 sin2 x+3 sin2 xdx      = 11+sin2 xdxDividing numerator and denominator by cos2 xI=sec2 x sec2 x+tan2 xdx      =sec2 x 1+tan2 x+tan2 xdx      = sec2 x 1+2 tan2dx      = sec2 x 1+2 tan x2dxLet 2 tan x=t2 sec2 x dx=dtsec2 x dx=dt2I=12  dt1+t2      =12 tan-1 t+C      =12 tan-1 2 tan x+C



Page No 19.117:

Question 1:

15+4 cos x dx

Answer:

Let I= 15+4 cos xdxPutting cos x= 1-tan2 x21+tan2 x2 I  = 15+41-tan2 x21+tan2 x2dx      = 1+tan2 x25 1+tan2 x2+41-tan2 x2dx    = sec2 x2 dx5+5 tan2 x2+4-4 tan2 x2    = sec2 x2 dxtan2 x2+9Let tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dtI=2  dtt2+32 =23tan-1 t3+C =23tan-1 tan x23+C

Page No 19.117:

Question 2:

15-4 sin x dx

Answer:

Let I= 15-4 sin xdxPutting sin x= 2 tan x21+tan2 x2I= 15-4×2 tan x21+tan2 x2dx    = 1+tan2 x251+tan2 x2-8 tan x2dx   = sec2 x2 5 tan2 x2-8 tan x2+5dxLet tan x2=t12 sec2x2dx=dtsec2 x2dx=2dtI=2  dt5t2-8t+5      =25 dtt2-85t+1      =25 dtt2-85t+452-452+1       =25  dtt-452-1625+1       =25  dtt-452+352      =25×53 tan-1 t-4535+C      =23 tan-1 5t-43+C       =23tan-1 5 tan x2-43+C

Page No 19.117:

Question 3:

11-2 sin x dx

Answer:

Let I =  11-2 sin xdxPutting sin x=2 tan x21+tan2 x2I  =11-2 ×2 tan x21+tan2 x2dx       = 1+tan2 x2 1+tan2 x2-4 tan x2dx      = sec2 x2tan2 x2-4 tan x2+1 dxLet tan x2=tsec2 x2×12dx=dtsec2 x2dx=2dtI=2 dtt2-4t+1     =2 dtt2-4t+4-4+1     =2  dtt-22-3      =2  dtt-22-32      =2×123ln t-2-3t-2+3+C     =13ln tan x2-2-3tan x2-2+3+C

Page No 19.117:

Question 4:

14 cos x-1 dx

Answer:

Let I = 14 cos x-1dxPutting cos x= 1-tan2 x21+tan2 x2I= 141-tan2 x21+tan2 x2-1dx    = 141-tan2 x2-1+tan2 x21+tan2 x2   = 1+tan2 x2dx4-4 tan2 x2-1-tan2 x2  = sec2 x2 dx3-5 tan2 x2Let tan x2=t12 sec2 x2dx=dt sec2 x2dx=2dtI=2  dt3-5 t2      =25  dt35-t2      =25  dt352-t2     =25×523ln 35+t35-t+C    =115ln 3+5 t3-5 t+C    =115ln 3+5 tan x23-5 tan x2+C

Page No 19.117:

Question 5:

11-sin x+cos x dx

Answer:

Let I= 11-sin x+cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2      = 11-2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx     = 1+tan2 x21+tan2 x2-2 tan x2+1-tan2 x2dx      = sec2 x22-2 tan x2dx     =12 sec2 x21-tan x2dxLet 1-tan x2=t-sec2 x2×12dx=dtsec2 x2dx=-2dtI=12  -2 dtt     =- dtt     =- ln t+C     =-ln 1-tan x2+C

Page No 19.117:

Question 6:

13+2 sin x+cos x dx

Answer:

Let I= 13+2 sin x+cos xdxPutting sin x=2 tan x21+tan2 x2and cos x=1-tan2 x21+tan2 x2I  = 13+2×2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx       = 1+tan2 x231+tan2 x2+4 tan x2+1-tan2 x2dx       = sec2 x23+3 tan2 x2+4 tan x2+1-tan2 x2 dx       = sec2 x22 tan2 x2+4 tan x2+4dx     =12 sec2 x2tan2 x2+2 tan x2+2dxLet tan x2=t sec2 x2×12 dx=dtsec2 x2dx=2dtI=12  2 dtt2+2 t+2      = dtt2+2t+1+1      = dtt+12+12      =tan-1 t+11+C     =tan-1 1+tan x2+C

Page No 19.117:

Question 7:

113+3 cos x+4 sin x dx

Answer:

Let I= 113+3 cos x+4 sin xdxPutting cos x =1-tan2 x21+tan2 x2 and sin x=2tan x21+tan2 x2I = 113+3 1-tan2 x21+tan2 x2+4×2tan x21+tan2 x2dx    = 1+tan2 x2131+tan2 x2+3-3 tan2 x2+8 tan x2 dx    = sec2 x2 13 tan2 x2-3 tan2 x2+16+8 tan x2dx    = sec2 x2 10 tan2 x2+8 tan x2+16dxLet tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dtI= 2 dt10t2+8t+16    = dt5t2+4t+8    =15  dtt2+45t+85    =15 dtt2+45t+252-252+85    =15 dtt+252-425+85    =15 dtt+252+-4+4025    =15 dtt+252+652    =15×56tan-1 t+2565+C    =16tan-1 5t+26+C    =16 tan-1 5 tan x2+26+C

Page No 19.117:

Question 8:

1cos x-sin x dx

Answer:

Let I= dxcos x-sin xPutting cos x=1-tan2 x21+tan2 x2 and sin x=2 tan x21+tan2 x2I= dx1-tan2 x21+tan2 x2-2 tan x21+tan2 x2     = sec2 x2dx1-tan2 x2-2 tan x2Let tan x2=t12 sec2 x2dx=dtsec2 x2dx=2 dtI= 2 dt1-t2-2t     = -2 dtt2+2t-1     = -2dtt2+2t+1-2     =- 2dtt+12-22     = 2dt22-t-12     =222ln 2+t+12-t-1+C     =12 ln 2+tan x2+12-tan x2-1+C

Page No 19.117:

Question 9:

1sin x+cos x dx

Answer:

Let I=1sin x+cos xdxPutting sin x =2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2       = 12 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx        = sec2 x21-tan2 x2+2 tan x2 dxLet tan x2=t 12sec2 x2dx=dtsec2 x2dx=2dtI=2 dt1-t2+2t      =-2  dtt2-2t-1      =-2  dtt2-2t+1-2      =2 dt22-t-12      =2×122ln 2+t-12-t+1+C      =12ln 2+tan x2-12-tan x2+1+C

Page No 19.117:

Question 10:

15-4 cos x dx

Answer:

Let I= 15-4 cos xdxPutting cos x=1-tan2 x21+tan2 x2  I= 15-4 1-tan2 x21+tan2 x2dx    = 1+tan2 x25 1+tan2 x2-4+4 tan2 x2dx    = sec2 x2 9 tan2 x2+1dxLet tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dtI=2dt9t2+1     =29dtt2+19     =29 dtt2+132     =29×3 tan-1 t13+C     =23 tan-1 3t+C     =23 tan-1 3 tan x2+C

Page No 19.117:

Question 11:

12+sin x+cos x dx

Answer:

Let I= 12+sin x+cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2I  = 12+2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx      = 1+tan2 x2 21+tan2 x2+2 tan x2+1-tan2 x2dx     = sec2 x2 2+2tan2 x2+2 tan x2+1-tan2 x2dx     = sec2 x2 tan2 x2+2 tan x2+3dxLet tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dt I=2 dtt2+2t+3     =2 dtt2+2t+1+2     =2 dtt+12+22     =2×12 tan-1 t+12+C      =2 tan-1 tan x2+12+C

Page No 19.117:

Question 12:

1sin x+3 cos x dx

Answer:

Let I= 1sin x+3 cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2I  = 12 tan x21+tan2 x2+31-tan2 x21+tan2 x2dx      = 1+tan2 x2 2 tan x2+3-3tan2 x2dx      =sec2 x2-3tan2 x2+2 tan x2+3dx

Let tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dt I=2 dt-3t2+2t+3=-23 dtt2-23t-1=-23dtt2-23t+132-132-1=-23 dtt-132-232=-23 ×1223log t-13-23t-13+23+C

=-12log t-33t+13+C=-12log 3t-33t+1+C=12log 3t+13t-3+C=12log 3tanx2+13tanx2-3+Cor, 12log 1+3tanx23-3tanx2+C

Page No 19.117:

Question 13:

13 sin x+cos x dx

Answer:

Let I= dx3 sin x+cos xPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2I  = 132 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx      = 1+tan2 x2 23 tan x2+1-tan2 x2dx      =sec2 x2-tan2 x2+23 tan x2+1dx

Let tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dt I=2 dt-t2+23t+1=-2 dtt2-23t-1=-2dtt2-23t+32-32-1=-2 dtt-32-22=-22×2log t-3-2t-3+2+C

=-12logtanx2-2-3tanx2+2-3+C=12logtanx2+2-3tanx2+2-3+C

Page No 19.117:

Question 14:

1sin x-3 cos x dx

Answer:

Let I= dxsin x-3 cos xPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2I  = 12 tan x21+tan2 x2-31-tan2 x21+tan2 x2dx      = 1+tan2 x2 2 tan x2-3+3tan2 x2dx      =sec2 x23tan2 x2+2 tan x2-3dx

Let tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dt I=2 dt3t2+2t-3=23 dtt2+23t-1=2dtt2+23t+132-1-132=2 dtt+132-232=22×23log t+13-23t+13+23+C
=32log tanx2-13tanx2+33+C=32log 3 tanx2-13 tanx2+3+C

Page No 19.117:

Question 15:

15+7 cos x+sin x dx

Answer:

Let I= 15+7 cos x+sin xdxPutting cos x=1-tan2 x21+tan2 x2 and sin x=2 tan x21+tan2 x2 I =  15+7 1-tan2 x21+tan2 x2+2 tan x21+tan2 x2 dx       = sec2 x251+tan2 x2+7-7 tan2 x2+2 tan x2dx      = sec2 x2-2 tan2 x2+2 tan x2+12dxLet tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dtI= 2 dt-2t2+2t+12      = dt-t2+t+6      = -dtt2-t-6      = -dtt2-t+122-122-6      = -dtt-122-14-6      = -dtt-122-522      = dt522-t-122      =12×52log 52+t-1252-t+12+C      =15log 2+t3-t+C      =15log 2+tan x23-tan x2+C



Page No 19.122:

Question 1:

11-cot x dx

Answer:

Let I= 11-cot xdx       =11-cos xsin xdx       =sin xsin x-cos xdx       =122 sin xsin x-cos x dx       =12sin x+cos x+sin x-cos xsin x-cos xdx       =12sin x+cos xsin x-cos xdx+12dxPutting sin x -cos x =tcos x+sin x dx=dt I=121tdt+12dx      =12 ln t+x2+C      =x2+12 ln sin x-cos x+C 

Page No 19.122:

Question 2:

11-tan x dx

Answer:

Let I= 11-tan xdx       =11-sin xcos xdx       =cos x cos x-sin xdx       =122 cos x cos x-sin xdx       =12cos x+sin x+cos x-sin xcos x-sin xdx       =12cos x+sin xcos x-sin xdx+12dxPutting cos x-sin x=t-sin x-cos xdx=dtsin x+cos xdx=-dt I=-12dtt+x2+C        =-12 ln cos x-sin x+x2+C      =x2-12 ln cos x-sin x+C

Page No 19.122:

Question 3:

3+2 cos x+4 sin x2 sin x+cos x+3 dx

Answer:

Let I=3+2 cos x +4 sin x 2 sin x+cos x+3dxLet 3+2 cos x+4 sin x=A 2 sin x+cos x+3 +B 2 cos x-sin x +C3+2 cos x+4 sin x=2A-B sin x+A+2B cos x+3A+C   

Comparing the coefficients of like terms
2A-B=4      ...  1A+2B=2      ...  (2)3A+C=3      ...  (3)

Multiplying eq (1) by 2 and adding it to eq (2) we get ,


4A-2B+A+2B=8+25A=10A=2

Putting value of A = 2 in  eq (1)

2×2-B=4B=0Putting value of A in eq (3) 3×2+C=3 C=-3

 I=2 2 sin x+cos x+3-32 sin x+cos x+3dx      =2dx-312 sin x+cos x+3dxSubstituting sin x=2 tan x21+tan2 x2 and  cos x =1-tan2 x21+tan2 x2 I=2dx-312×2 tan x21+tan2 x2+1-tan2 x21+tan2 x2+3dx      =2dx-31+tan2 x2 4 tan x2+1-tan2 x2+3 1+tan2 x2dx      =2dx-3sec2 x22 tan2 x2+4 tan x2+4 dx      =2dx-32sec2 x2 tan2 x2+2 tan x2+2dxPutting tan x2=t12 sec2 x2 dx=dtsec2 x2 dx=2dt I=2dx-322t2+2t+2 dt      =2dx-31t2+2t+1+1dt      =2dx-31t+12+12dt      =2x-31 tan-1 t+11+C      =2x-3 tan-1 tan x2+1+C                   t= tan x2

Page No 19.122:

Question 4:

1p+q tan x dx

Answer:

Let I=dxp+q tan x       =1p+q sin xcos xdx       =cos x q sin x+p cos xdxLet cos x=A q sin x+p cos x+B q cos x-p sin xcos x=Ap+Bq cos x+Aq-Bp sin x

Comparing coefficients of like terms

Ap+Bq=1     ...   1Aq-Bp=0     ...   2

Multiplying eq (1) by p and eq (2) by q and then adding

Ap2+Bpq=pAq2-Bpq=0A=pp2q2

Putting value of A in eq (1)

p2p2+q2+Bq=1Bq=1-p2p2+q2Bq=p2+q2-p2p2+q2B=qp2+q2 I=pp2+q2×q sin x+p cos xq sin x +p cos x+qp2+q2×q cos x-p sin xq sin x+p cos xdx      =pp2+q2dx+qp2+q2q cos x-p sin xq sin x +p cos xdxPutting q sin x+p cos x=tq cos x-p sin x dx=dt I=pp2+q2dx+qp2+q21tdt      =pp2+q2 x+qp2+q2 ln q sin x+p cos x+C         

Page No 19.122:

Question 5:

5 cos x+62 cos x+sin x+3 dx

Answer:

Let I=5 cos x+62 cos x+sin x+3dx& let 5 cos x+6=A 2 cosx+sin x+3+B-2 sin x+cos x+C       ....(1) 5 cos x+6=A-2B sin x+2A+B cos x +3A+C

Comparing coefficients of like terms

A-2B=0    ...  22A+B=5    ...  (3)3A+C=6    ...  (4)

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10
A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

By putting the values of A,B and C in eq (1) we get , I=2 2 cos x+sin x+3+-2 sin x+cos x2 cos x+sin x+3dx      =2dx+ -2 sin x+cos x2 cos x+sin x+3dxPutting 2 cos x+sin x+3=t-2 sin x+cos xdx=dt I=2dx+1tdt      =2x+ln 2 cos x+sin x+3+C

Page No 19.122:

Question 6:

2 sin x+3 cos x3 sin x+4 cos x dx

Answer:

Let I=2 sin x+3 cos x3 sin x+4 cos xdx& let 2 sin x+3 cos x=A 3 sin x+4 cos x+B 3 cos x-4 sin x      ...(1)2 sin x+3 cos x=3A-4B sin x+4A+3B cos x

By comparing the coefficients of like terms we get,

3A-4B=2   ...  24A-3B=3   ...  3

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

9A-12B+16A+12B=6+1225A=18A=1825Putting value of A=1825 in eq 2 we get,3×1825-4B=25425-2=4B425×4=BB=125
Thus, substituting the values of A,B and C in eq (1) we get ,

 I =18253 sin x+4 cos x+125 3 cos x-4 sin x3 sin x+4 cos xdx   =1825dx+1253 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t3 cos x-4 sin x dx=dt I=1825dx+1251tdt      =18x25+125 ln t+C      =18x25+125 ln 3 sin x+4 cos x+C

Page No 19.122:

Question 7:

13+4 cot x dx

Answer:

Let I=13+4 cot xdx       =13+4 cos xsin xdx       =sin x 3 sin x+4 cos xdxLet sin x=A3 sin x+4 cos x+B 3 cos x-4 sin x         ...(1)sin x=3A-4B sin x+4A+3B cos xBy comparing the coefficients of both sides we get ,3A-4B=1   ...   24A+3B=0   ...   3

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

9A-12B+16A+12B=3+025A=3A=325Putting value of A in eq 3 we get,4×325+3B=03B=-1225B=-425
Thus, by substituting the value of A and B in eq (1) we getI=3253 sin x+4 cos x-4253 cos x-4 sin x3 sin x+4 cos xdx =325dx-4253 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t3 cos x-4 sin xdx=dtI=325dx-425dtt     =325x-425 ln t+C     =3x25-425 ln 3 sin x+4 cos x+C

Page No 19.122:

Question 8:

2 tan x+33 tan x+4 dx

Answer:

Let I=2 tan x+33 tan x+4dx      =2 sin xcos x+33 sin xcos x+4dx      =2 sin x+3 cos x3 sin x+4 cos xdxLet 2 sin x+3 cos x=A 3 sin x+4 cos x+B 3 cos x-4 sin x             ....(1) 2 sin x+3 cos x=3A-4B sin x+4A+3B cos xEquating the coefficients of like terms3A-4B=2   ...   24A+3B=3   ...   3

Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get

9A-12B=616A+12B=12        25A=18A=1825Putting value of A in eq 2 we get,B=125
Thus, by substituting the values of A and B in eq (1) we get,I=  1825 3 sin x+4 cos x+1253 cos x-4 sin x3 sin x+4 cos xdx =1825dx+1253 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t3 cos x-4 sin xdx=dt I=1825x+1251tdt      =18x25+125 ln t+C      =18x25+125 ln 3 sin x+4 cosx+C

Page No 19.122:

Question 9:

14+3 tan x dx

Answer:

Let I=dx4 +3 tan x=dx4+3 sin xcos x=cos x dx4 cos x+3 sin xConsider,cos x=A 4 cos x+3 sin x+Bddx4 cos x+3 sin xcos x=A 4 cos x+3 sin x+B -4 sin x+3 cos xcos x=4A+3B cos x+3A-4B sin xEquating the coefficients of like terms4A+3B=1           .....13A-4B=0           .....2
Solving (1) and (2), we get
A=425 and B=325
4254 cos x+3 sin x+-4 sin x+3 cos x3254 cos x+3 sin xdx=425dx+325-4 sin x+3 cos x4 cos x+3 sin xdxlet 4 cos x+3 sin x=t-4 sin x+3 cos xdx=dtThen,I=425dx+325dtt=4x25+325 log t+C=4x25+325 log 4 cos x+3 sin x+C

Page No 19.122:

Question 10:

8 cot x+13 cot x+2 dx

Answer:

Let I=8 cot x+13 cot x+2dx       =8 cos xsin x+13 cos xsin x+2dx      =8 cos x+sin x3 cos x+2 sin xdxNow, let 8 cos x+sin x=A 3 cos x+2 sin x+B -3 sin x+2 cos x             ...(1) 8 cos x+sin x=3A cos x+2A sin x-3B sin x+2B cos x    8 cos x+sin x=3A+2B cos x+2A-3B sin x Equating the coefficients of like terms we get, 2A-3B=1   ...   23A+2B=8   ...   3

Solving eq (2) and  eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,


I=2 3 cos x+2 sin x+1-3 sin x+2 cos x3 cos x+2 sin xdx =23 cos x+2 sin x3 cos x+2 sin xdx+-3 sin x+2 cos x3 cos x+2 sin xdx =2dx+-3 sin x+2 cos x3 cos x+2 sin xdxPutting 3 cos x+2 sin x=t-3 sin x+2 cos xdx=dt I=2dx+1tdt      =2x+ln t+C      =2x+ln 3 cos x+2 sin x+C

Page No 19.122:

Question 11:

4 sin x+5 cos x5 sin x+4 cos x dx

Answer:

Let I=4 sin x+5 cos x5 sin x+4 cos xdx& let 4 sin x+5 cos x=A 5 sin x+4 cos x+B 5 cos x-4 sin x       ...(1) 4 sin x+5 cos x=5A-4B sin x+4A+5B cos xBy equating the coefficients of like terms we get, 5A-4B=4   ...   24A+5B=5   ...   3

By solving eq (2) and eq (3) we get,
A=4041, B=941Thus, by substituting the values of A and B in eq (1) , we getI=40415 sinx +4 cos x+9415 cos x-4 sin x5 sin x+4 cos xdx =4041dx+9415 cos x-4 sin x5 sin x+4 cos xdxPutting 5 sin x+4 cos x=t5 cos x-4 sin xdx=dtI=4041x+9411tdt     =40 41x+941 ln t+C     =4041x+941 ln 5 sin x+4 cos x+C



Page No 19.133:

Question 1:

x cos x dx

Answer:

 x cosx dxTaking x as the first function and cos x as the second function.= xcosx dx-ddxxcosx dxdx= x sinx-sinx dx= x sinx+cosx+C

Page No 19.133:

Question 2:

log x+1 dx

Answer:

 log x+1dx= 1. log x+1dxTaking log x+1 as the first function and 1 as the second function.= log x+11 dx-ddxlogx+11 dxdx= x log x+1-xx+1dx= x log x+1-x+1x+1-1x+1dx= x log x+1-x+log x+1+C

Page No 19.133:

Question 3:

x3 log x dx

Answer:

x3 logx dxTaking log x as the first function and x3 as the second function.= logxx3dx-ddxlogxx3dxdx= log xx44-1xx44dx= x4x logx-x416+C

Page No 19.133:

Question 4:

xex dx

Answer:

xex dxTaking x as the first function and ex as the second function.=xexdx-ddxxex dxdx= xex-1exdx= xex-ex+C= x-1ex+C

Page No 19.133:

Question 5:

xe2x dx

Answer:

xe2x dxTaking x as the first function and e2x as the second function .= xe2xdx-ddxxe2x dxdx= x e2x2-e2x2dx= x2e2x-e2x4+C= e2xx2-14+C

Page No 19.133:

Question 6:

x2 e-x dx

Answer:

x2 e-x dxTaking x2 as the first function and e-x as the second function.= x2e-xdx-ddxx2e-xdxdx= -x2 e-x-2xe-x-1dx= -x2 e-x+2x e-x dx= - x2 e-x+2-x e-x+e-x dx= -x2 e-x+2-x e-x -e-x+C= -e-xx2+2x+2+C

Page No 19.133:

Question 7:

x2 cos x dx

Answer:

x2 cosx dxTaking x2 as the first function and cos x as the second function.= x2cosx dx-ddxx2cosx dxdx= x2sinx-2x sinx dx= x2sinx-2xsinx-ddxxsinx dxdx= x2sinx-2-xcosx+cosx dx= x2sinx+2x cosx-2 sinx+C

Page No 19.133:

Question 8:

x2 cos 2x dx

Answer:

x2 cos 2x dxTaking x2 as the first function and cos 2x as the second function.= x2cos 2x dx-2xcos 2x dxdx= x2 sin 2x2-2x sin 2x2dx= x22sin 2x-x sin 2x dx= x22sin 2x-xsin2x-sin 2x dxdx= x22sin 2x--x cos 2x2+cos 2x2dx= x22sin 2x+x cos 2x2-sin 2x4+C

Page No 19.133:

Question 9:

x sin 2x dx

Answer:

x sin 2x dx Taking x as the first function and sin 2x as the second function.= xsin2x dx-ddxxsin 2x dxdx= -x cos 2x2+cos 2x2dx= -x cos 2x2+sin 2x4+C

Page No 19.133:

Question 10:

log log xx dx

Answer:

log log xxdxTaking  log log x as the first function and 1x as the second function.= log log x1xdx-ddx log log x1xdxdx= log x.log log x-1x log xlog xdx= log x.log log x-1xdx= log x.log log x-log x+ C 

= log xlog log x-1+C

Page No 19.133:

Question 11:

x2 cos x dx

Answer:

x2 cosx dxTaking x2 as the first function and cos x as the second function.= x2cosx dx-ddxx2cosx dxdx= x2sinx-2x sinx dx= x2sinx-2xsinx-ddxxsin x dxdx= x2sinx+2xcosx-2cos x dx= x2sinx+2x cosx-2 sinx+C

Page No 19.133:

Question 12:

x cosec2 x dx

Answer:

x cosec2x dx Taking x as the first function and cosec2 x as the second function.= xcosec2x dx-ddxxcosec2x dxdx= -x cotx+cotx dx= -x cotx+log sinx+c

Page No 19.133:

Question 13:

x cos2 x dx

Answer:

x cos2x dxTaking x as the first function and cos2x as the second function.= x1+cos 2x 2dx-ddxx1+cos 2x 2dxdx= x2x+sin2x2-12x+sin2x2dx= x2x+sin2x2-x24-cos2x8+C= x22+x sin2x2-x24+cos2x8+C= x24+x sin2x2+cos2x8+C

Page No 19.133:

Question 14:

xn·log x dx

Answer:

xn logx dxTaking log x as the first function and xn as the second function.= logxxndx -ddx logxxn dxdx= logxxn+1n+1-1xxn+1n+1dx= logxxn+1n+1-xnn+1dx= logxxn+1n+1-xn+1n+12+C

Page No 19.133:

Question 15:

log xxn dx

Answer:

1xn logx dxTaking log x as the first function and 1xnas the second function.= logx1xndx-ddxlogx1xndxdx= logxx-n+1-n+1-1xx-n+1-n+1dx= logxx-n+1-n+1-x-n-n+1dx= logxx-n+1-n+1-x-n+1-n+12+C=logxx1-n1-n-x1-n1-n2+C

Page No 19.133:

Question 16:

x2 sin2 x dx

Answer:

x2 sin2x dxTaking x2 as the first function and sin2 x as the second function.= x21-cos2x2-ddxx21-cos2x2dxdx= x22x-sin2x2-2xx-sin2x22dx= x22x-sin2x2-x2dx+xsin2x2dx           Here, taking x as the first function and sin 2x as the second function.=x32-x2sin2x4-x33+12xsin 2x-ddxxsin 2x dxdx= x32-x2sin2x4-x33+12-xcos2x2+cos2x dx4= x36-x2sin2x4-x cos2x4+sin2x8+C

Page No 19.133:

Question 17:

2x3 ex2 dx

Answer:

    2x3·ex2dx=x2·ex2·2x dxLet x2=t2x dx=dt=tI·eIIt dt=t·et-1·et dt=t et-et+C=x2 ex2-ex2+C=ex2x2-1+C

Page No 19.133:

Question 18:

x3 cos x2 dx

Answer:

x3 cosx2 dx Let  x2=t                                       2x=dtdxdx=dt2x= 12t cost dtTaking t as the first function and cos t as the second function .= 12tsint-sint dt= 12tsint+cost          ...(1) Substituting the value of t in eq (1) = x2sinx22+cosx22+c

Page No 19.133:

Question 19:

x sin x cos x dx

Answer:

    xsin x·cos x dx=12x2 sin x cos x dx=12x·sin 2x dxTaking x as the first function and sin 2x as the second function . =12xsin 2x dx -ddxxsin 2x dxdx=12x×-cos 2x2-1·-cos 2x2dx=12-x cos 2x2+sin 2x4+C=-x cos 2x4+sin 2x8+C

Page No 19.133:

Question 20:

sin x log cos x dx

Answer:

Let I=sin x·log cos x dxLet cos x =t-sin x dx=dtsin x dx=-dt  I=-log t dt      =-1·log t dtTaking log t as the first function and 1 as the second function .     =log t1 dt-ddt log t1dtdt      =-log t ·t-1t×t dt     =-log t·t-t+C     =-tlog t-1+C       ....(1) Substituting the value of t in eq (1)     =-cos xlog cos x-1+C     =cos x 1-log cos x+C

Page No 19.133:

Question 21:

log x2·x dx

Answer:

log x2 x ·dxTaking log x2 as the first function and x as the second function . =log x2x dx-ddxlog x2xdxdx=log x2·x22-2 log xx×x22 dx=log x2×x22-xII log xI dx=log x2×x22-logx x dx -ddxlog xx dxdx=log x2×x22-log x·x22-1x×x22dx=log x2×x22-log x·x22+x24+C=x22log x2-log x+12 +C

Page No 19.133:

Question 22:

ex dx

Answer:

Let I =ex dx         =x·exxdxLet x=t12xdx=dtdxx=2 dt I=2t·et dtTaking t as the first function and et as the second function .        =2tetdt-ddttetdtdt         =2t·et-1·et dt+C        ...(1)Substituting the value of t in eq(1)       =2x ex-ex+C      =2exx-1+C

Page No 19.133:

Question 23:

log x+2x+22 dx

Answer:

Let I=log x+2 dxx+22Let log x+2=tx+2=et1x+2dx=dtI=tetdt     =t e-t dtTaking t as the first function and e-t as the second function.     =te-t-ddtte-2tdtdt    =t×e-t-1-1·e-t dt     =-t e-t+e-t-1+C     =-e-tt+1+C     =-t+1et+C        ...(1)Substituting the value of t in eq (1)      =-log x+2+1x+2+C     =-log x+2 x+2-1x+2+C

Page No 19.133:

Question 24:

x+sin x1+cos x dx

Answer:

x+sin x1+cos xdx=x1+cos x+sin x1+cos xdx=x2 cos2 x2+2 sin x2 cos x22 cos2 x2dx=12xI·sec2II x2dx+tan x2dx=12x·tan x212-1×2 tan x2dx+log sec x212+C=x tan x2-log sec x212+log sec x212+C=x tan x2+C

Page No 19.133:

Question 25:

log10 x dx

Answer:

     log10 x dx=log xlog 10dx=1log 101·log x dxTaking log x as the first function and 1 as the second function=1log 10logx 1 dx-ddxlog x1 dxdx=1log 10log x·x-1x·x dx=1log 10x log x-x+C=1log 10xlog x-1+C

Page No 19.133:

Question 26:

cosx dx

Answer:

Let I=cos x dx       =x· cos xxdxLet x=t12xdx=dtdxx=2dtI=2t·cos t·dtTaking t as the first function and cos t as the second function .     =2 t·sin t-1·sin t dt     =2 t·sin t+cos t+C    ....(1) Substituting the value of t in eq (1)     =2 x·sin x+cos x+C

Page No 19.133:

Question 27:

Evaluate the following integrals:

x cos-1x1-x2dx

Answer:

Let I=x cos-1x1-x2dxLet the first function be cos-1x and second function be x1-x2.First we find the integral of the second function, i.e., x1-x2dx.Put t=1-x2. Then dt=-2xdxTherefore,x1-x2dx=-121tdt                     =-t                     =-1-x2Hence, using integration by parts, we getx cos-1x1-x2dx=cos-1xx1-x2dx-dcos-1xdxx1-x2dxdx                      =cos-1x-1-x2--11-x2-1-x2dx                      =-1-x2 cos-1x-x+cHence, x cos-1x1-x2dx=-1-x2 cos-1x-x+c

Page No 19.133:

Question 28:

Evaluate the following integrals:

logxx+12dx

Answer:

Let I=logxx+12dxLet the first function be logx and second function be 1x+12.First we find the integral of the second function, i.e., 1x+12dx.Put t=x+1. Then dt=dxTherefore,1x+12dx=t-2 dt                    =-1t                    =-11+xHence, using integration by parts, we getlogxx+12dx=logx1x+12dx-dlogxdx1x+12dxdx                    =logx-11+x-1x-11+xdx                    =-logx1+x+1x2+xdx                    =-logx1+x+1x2+x+14-14dx                    =-logx1+x+1x+122-122dx                    =-logx1+x+12×12logx+12-12x+12+12+c                    =-logx1+x+logxx+1+cHence, logxx+12dx=-logx1+x+logxx+1+c

Page No 19.133:

Question 29:

cosec3 x dx

Answer:

Let I =cosec3x dx         =cosec2x·cosec x dx        =cosec2x·1+cot2x dxLet cot x=t-cosec2x dx=dt I=-1+t2dt      =-t21+t2-122 log t+1+t2+C      ...(1)Substituting the value of t in eq (1)      =-cot x2·cosec x-12 log cot x+cosec x+C      =-12cosec x cot x-12 log cos xsin x+1sin x+C      =-12 cosec x cot x-12 log 2 cos2 x22 sin x2 cos x2+C      =-12 cosec x cot x-12 log cot x2+C      =-12 cosec x cot x+12 log tan x2+C         log cot x2=log 1tan x2 -log tan x2 

Page No 19.133:

Question 30:

sec-1 x dx

Answer:

 1II.sec-1 xI dx=sec-1 x1 dx-ddxsec-1 x1 dxdx=sec-1 x.x- 1x 1-x×12x×x dx= x sec-1 x-12  1-x-12 dx=x sec-1 x-12 1-x-12+1-12+1 -1+C=x sec-1 x+1-x12+C



Page No 19.134:

Question 31:

sin-1 x dx

Answer:

Let I= sin-1 x dx= x.sin-1 xxdxLet x=t12xdx=dtdxx=2dtI= tII.sin-1 tI dt     =sin-1 tt dt-ddtsin-1 ttdtdt     =2 sin-1 t.t22- 11-t2×t22dt     =sin-1 t.t2- t21-t2dt     =sin-1 t.t2+1-t2-11-t2dt     =sin-1 t.t2+ 1-t2 dt- dt1-t2     =sin-1 t.t2+t21-t2+12sin-1 t-sin-1 t+C     =sin-1 t.t2+t21-t2-12sin-1t+C     =x.sin-1 x+x2 1-x-12sin-1 x+C          x=t     =2x-1 sin-1 x2+x-x22+C

Page No 19.134:

Question 32:

x tan2 x dx

Answer:

 x tan2 x dx
= â€‹∫ x (sec2 x – 1) dx
= xI. sec2 xII dx- x dx=xsec2x-ddxxsec2x dxdx-x22+C1=x.tan x-1.tan x dx-x22+C1=x tan x-log sec x-x22+C1+C2=x tan x-log sec x-x22+C        where C=C1+C2

Page No 19.134:

Question 33:

xsec 2x-1sec 2x+1 dx

Answer:

 xsec 2x-1sec 2x+1dx= x 1cos 2x-11cos 2x+1dx= x 1-cos 2x1+cos 2xdx= x 2 sin2 x2 cos2 xdx           1-cos 2x=2sin2 x and 1+cos 2x=2 cos2x= x. tan2 x dx= x.sec2 x-1 dx= xI.sec2 xII dx- x dx=xsec2x dx-ddxxsec 2x dxdx-x22+C1=x tan x-1. tan x dx-x22+C1=x tan x-log sec x-x22+C2+C1=x tan x-log sec x-x22+C         where C=C1+C2

Page No 19.134:

Question 34:

x+1 ex log xex dx

Answer:

 x+1ex .log x ex dxLet x ex=tx.ex+1.exdx=dt x+1ex .log x ex dx=1II.log tIdt                                             =log t1 dt-ddtlog t-1dtdt                                             =log t×t-1t×t dt                                             =t log t-t+C           ...(1)Substituting the value of t in eq (1) x+1ex .log x ex dx=x ex.log x ex-x ex+C                                             =x exlogx ex-1+C

Page No 19.134:

Question 35:

sin-1 3x-4x3 dx

Answer:

sin–1 (3x – 4x3)dx
Let x = sin θ
dx = cos​ θ.d​θ
& θ = sin–1 x
sin–1 (3x – 4x3)dx = sin–1 (3 sin â€‹θ – 4 sin3 â€‹θ) . cos â€‹θ d​θ
                                = ∫ sin–1 (sin 3​θ) . cos â€‹θ d​θ
                                =3 θI.cos θII dθ=3θcos θdθ-ddθθ-cos θ dθdθ =3θ.sin θ-1.sin θ dθ=3θ.sin θ+cos θ+C=3θ.sin θ+1-sin2 θ+C=3sin-1 x.x+1-x2+C            θ= sin-1x

Page No 19.134:

Question 36:

sin-12x1+x2 dx

Answer:

 sin-1 2x1+x2dxLet x=tan θdx=sec2 θ dθ sin-1 2x1+x2dx= sin-1 2 tan θ1+tan2 θ. sec2 θ dθ                                      = sin-1 sin 2θ.sec2 θ dθ                                      = 2θ sec2 θ dθ                                      =2 θI sec2 θII dθ                                      =2θsec2θ dθ-ddθθsec2θ dθdθ                                      =2θ.tan θ-1.tan θ dθ                                      =2θ tan θ-log sec θ+C                                      =2θ tan θ-log 1+tan2 θ12+C                                      =2tan-1 x×x-log 1+x212+C                                      =2 x tan-1 x-2×12log 1+x2+C                                     =2 x tan-1 x-log 1+x2+C

Page No 19.134:

Question 37:

tan-13x-x31-3x2 dx

Answer:

Let I= tan-1 3x-x31-3x2 dx=3 tan-1 x dx=3tan-1 x×1 dx=3 tan-1 x×x-11+x2×x dx=3x tan-1 x-3x1+x2 dxlet 1+x2=t2x dx=dtThen,I=3x tan-1x-32dtt=3x tan-1x-32 log t+C=3x tan-1x-32 log 1+x2+C

Page No 19.134:

Question 38:

x2 sin-1 x dx

Answer:

 x2II.sin-1Ix dx=sin-1x x2 dx-ddxsin-1xx2 dxdx=sin-1x.x33-11-x2 x33dxLet 1-x2=tx2=1-t-2x dx=dtx dx=-dt2 x2.sin-1x dx=sin-1 x.x33-13 x2.x1-x2dx                            =sin-1 x.x33-16 1-ttdt                            =sin-1 x.x33+16t-12 dt-16t12 dt                            =sin-1 x.x33+16×2t-16×23t32+C                            =sin-1 x.x33+1-x23-191-x232+C          1-x2=t

Page No 19.134:

Question 39:

sin-1 xx2 dx

Answer:

Let I= sin-1 xx2 dxPutting x=sin θθ=sin-1 x & dx=cos θ dθ I= θ.cos θ sin2 θdθ      = θ.cos θsin θ×1sin θ dθ      = θI.cosec θII cot θ dθ      =θcosec θ cot θ dθ-ddθθcosec θ cot θ dθdθ      =θ -cosec θ-1.-cosec θ dθ      =-θ cosec θ+ cosec θ dθ      =-θ cosec θ+ln  cosec θ-cot θ+C      =-θsin θ+ln 1-cos θsin θ+C      =-θsin θ+ln 1-1-sin2 θsin θ+C      =-sin-1 xx+ln 1-1-x2x+C          θ=sin-1 x

Page No 19.134:

Question 40:

x2 tan-1 x1+x2 dx

Answer:

Let I= x2 tan-1 x1+x2dx        = x2+1-1x2+1tan-1 x dx       = 1-1x2+1tan-1 x dx       = 1II.tan-1I x dx- tan-1 xx2+1dx       =tan-1 x1 dx-ddxtan-1x1 dxdx- tan-1 xx2+1dx       =tan-1x×x-x1+x2dx-tan-1 xx2+1dxPutting  x2+1=t in the first integral and tan-1 x=p in the second integral2x dx=dt  and 11+x2dx=dpx dx=dt2and 11+x2dx=dp I=tan-1 x.x-12 dtt-p.dp       =x tan-1 x-12ln t-p22+C       =x tan-1 x-12ln 1+x2-tan-1 x22+C           t=x2+1 and p=tan-1x

Page No 19.134:

Question 41:

cos-1 4x3-3x dx

Answer:

 cos-1 4x3-3xdxLet x=cos θ θ=cos-1 x& dx=-sin θ dθ cos-1 4x3-3xdx= cos-1 4 cos3 θ-3 cos θ.-sin θdθ                                        = cos-1 cos 3θ.-sin θdθ                    cos 3θ=4 cos3 θ-3 cos θ                                        =-3  θI sin θII dθ                                        =θsin θ dθ-ddθθsin θ dθdθ                                        =3 θ -cos θ-1.-cos θdθ                                        =3θ cos θ-3 sin θ+C                                             =3 cos-1 x.x-31-x2+C              x=cos θ

Page No 19.134:

Question 42:

cos-1 1-x21+x2 dx

Answer:

Let I= cos-1 1-x21+x2dx        =2  1II. tan-1I x dx                   cos -11-x21+x2=2 tan-1x           =2tan-1x1 dx-ddxtan-1x1 dxdx        =2tan-1 x. x-11+x2×x dx        =2 x tan-1 x-2x1+x2dxPutting 1+x2=t2x dx=dt I=2x tan-1 x- dtt       =2x tan-1 x-ln t+C       =2x tan-1 x-ln 1+x2+C                  t=1+x2

Page No 19.134:

Question 43:

tan-1 2x1-x2 dx

Answer:

Let I= tan-1 2x1-x2dx       =2 1II.tan-1 xI dx      =2 tan-1x1 dx-ddxtan-1x1 dxdx       =2tan-1 x.x-11+x2×x dx       =2 tan-1 x.x- 2x1+x2dxPutting 1+x2=t2x dx=dt I=2x tan-1 x- dtt      =2x tan-1 x-ln t+C      =2x tan-1 x-ln 1+x2+C         t= 1+x2

Page No 19.134:

Question 44:

x+1 log x dx

Answer:

 x+1II.log xI dx=log xx+1dx-ddxlog xx+1dxdx=log xx22+x- 1xx22+xdx=log xx22+x- x2+1dx=log xx22+x-x24+x+C

Page No 19.134:

Question 45:

x2 tan-1 x dx

Answer:

Let I= x2II.tan-1 xI dx        =tan-1xx2dx-ddxtan-1xx2dxdx        =tan-1 x×x33- 11+x2×x33 dx        =tan-1 x.x33-13 x2. x1+x2dxLet 1+x2=t2x dx=dtx dx=dt2I=tan-1 x.x33-16 t-1t.dt      =tan-1 x.x33-16 dt+16 dtt      =tan-1x.x33-t6+16log t+C      =tan-1 x.x33-1+x26+16log 1+x2+C      =tan-1 x.x33-x26+16log 1+x2-16+C      =tan-1 x.x33-x26+16log 1+x2+C' where C'=C-16

Page No 19.134:

Question 46:

elog x+sin x cos x dx

Answer:

 elog x+sin x cos x dx= x+sin xcos x dx           elog x=x = x cos x+sin x cos x dx= x cos x dx+12 2 sin x cos x dx= xI cos xII dx+12 sin 2x dx=xcos x dx-ddxxcos x dxdx+12 sin 2x dx=x sin x-1.sin x dx+12-cos 2x2+C=x sin x--cos x-14cos 2x+C=x sin x+cos x-141-2 sin2 x+C=x sin x+cos x+sin2 x2-14+C=x sin x+cos x+sin2 x2+C'          where C'=C-14

Page No 19.134:

Question 47:

x tan-1 x1+x23/2 dx

Answer:

Let I= x tan-1 x1+x232dxPutting x=tan θdx=sec2 θ dθ& θ=tan-1 x I= tan θ.θ.sec2 θ dθ1+tan2 θ32     = θ.tan θ sec2 θ dθsec2 θ32     = θ tan θ.sec2 θ dθsec3 θ     = θ.tan θsec θ dθ     = θI.sin θII dθ    =θsin θ dθ-ddθθsin dθdθ     =θ -cos θ-1.-cos θ dθ     =-θ cos θ+ sin θ+C     =-θsec θ+1cosec θ+C     =-θ1+tan2 θ+11+cot2 θ+C     =-θ1+tan2 θ+tan θtan2 θ+1+C     =-tan-1 x1+x2+xx2+1+C

Page No 19.134:

Question 48:

tan-1 x dx

Answer:

Let I= tan-1 x dx       = x.tan-1 x dxxLet x=t12xdx=dt=dxx=2dtI=2 tII.tan-1 tI dt     =2 tan-1tt dt-ddttan-1tt dtdt     =2 tan-1 t.t22- 11+t2.t22dt     =tan-1 t.t2- t21+t2 dt     =tan-1 t.t2- 1+t2-11+t2dt     =tan-1 t.t2- dt+dt1+t2     =tan-1 t.t2-t+tan-1 t+C         x=t     =tan-1 x.x-x+tan-1 x+C     =x+1 tan-1 x-x+C

Page No 19.134:

Question 49:

x3 tan-1 x dx

Answer:

 x3II.tan-1 xI dx=tan-1x x3 dx-ddxtan-1xx3 dxdx=tan-1 x.x44-11+x2×x44dx=tan-1 x.x44-14 x4 dxx2+1=tan-1 x.x44-14 x4-1+1x2+1dx=tan-1 x.x44-14 x4-1x2+1dx-14 1x2+1dx=tan-1 x.x44-14x2-1 x2+1x2+1dx-14 1x2+1dx=tan-1x.x44-14 x2-1dx-14 tan-1 x+C=tan-1 x.x44-14x33-x-14tan-1 x+C=x4-14 tan-1x -112x3-3x+C

Page No 19.134:

Question 50:

x sin x cos 2x dx

Answer:

 x. cos 2x sin x dx=12  x 2 cos 2x sin x dx              2 cos A sin B=sin A+B-sin A-B=12  x sin 3x-sin x dx=12  x.sin 3x dx-12  x sin x dx=12xsin 3x dx-ddxxsin 3x dxdx-12xsin x dx-ddxxsinx dxdx=12 x.-cos 3x3- 1.-cos 3x3dx-12 x.-cos x- 1.-cos x dx=12x.-cos 3x3+19sin 3x -12x.-cos x+sin x=-x cos 3x6+sin 3x18+x cos x2-sin x2+C

Page No 19.134:

Question 51:

tan-1 x2 x dx

Answer:

Let  I = (tan–1 x2) x dx
Putting x2 = t
⇒​ 2x dx = dt
x dx=dt2 I=12 1II.tan-1 tI.dt      =12tan-1t1 dt-ddttan-1t1 dtdt      =12 tan-1 t. t- t1+t2dtNow putting 1+t2=p2t dt=dpt dt=dp2 I=12t. tan-1 t-12 t dt1+t2      =t.tan-1 t2-12x2  dpp      =t.tan-1 t2-14ln p+C      =x2.tan-1 x22-14 ln 1+x4+C        p=1+t2 

Page No 19.134:

Question 52:

x sin-1 x1-x2 dx

Answer:

 x.sin-1 x1-x2dxLet sin-1 x=θx=sin θdx=cos θ dθ x.sin-1 x1-x2dx= sin θ.θ1-sin2 θ.cos θ dθ                               = sin θ.θcos θ.cos θ dθ                               = θI.sin θII dθ                               =θsin θ dθ-ddθθsin θ dθdθ                               =θ-cos θ- 1.-cos θ dθ                               =-θ cos θ+sin θ+C                               =-θ 1-sin2 θ+sin θ+C                               =-sin-1 x 1-x2+x+C         sin-1 x=θ

Page No 19.134:

Question 53:

sin3 x dx

Answer:

Let, I=sin3x dx    .....1Consider, x=t         .....2Differentiating both sides we get,12xdx=dtdx=2x dtdx=2t dtTherefore, 1 becomes,I=sin3t 2t dt  =2tsin3t dt  =2t 3sint-sin3t4 dt             Since, sin3A=3sinA-4sin3A  =32t sint dt-12t sin3t dt  =32t sint dt- dtdtsint dtdt-12t sin3t dt- dtdtsin3t dtdt  =32-t cost +cost dt-12-t cos3t 3+13cos3t dt =32-t cost +sint -12-t cos3t 3+19sin3t +C =-32t cost+32sint+16t cos3t -118sin3t+C =-32xcosx+32sinx+16xcos3x-118sin3x+C     

Note: The answer in indefinite integration may vary depending on the integral constant.

Page No 19.134:

Question 54:

x sin3 x dx

Answer:

Let I =x sin3x dx
sin (3A) = 3 sin A – 4 sin3 A
sin3 A=143 sin A-sin 3AI=14 x.3 sin x-sin 3xdx     =34 xI.sin xII dx-14xI.sin 3xIIdx     =34x-cos x-1.-cos xdx-14x-cos 3x3-1.-cos 3x3dx     =-3x cos x4+34sin x+x cos 3x12-136sin 3x+C

Page No 19.134:

Question 55:

cos3 x dx

Answer:

Let, I=cos3x dx    .....1Consider, x=t         .....2Differentiating both sides we get,12xdx=dtdx=2x dtdx=2t dtTherefore, 1 becomes,I=cos3t 2t dt  =2t cos3t dt  =2t 3cost+cos3t4 dt             Since, cos3A=4cos3A-3cosA  =32t cost dt+12t cos3t dt  =32t cost dt- dtdtcost dtdt+12t cos3t dt- dtdtcos3t dtdt  =32t sint -sint dt+12t sin3t 3-13sin3t dt =32t sint+cost +12t sin3t 3+19cos3t +C =32t sint+32cost+16t sin3t +118cos3t+C=32xsinx+32cosx+16xsin3x+118cos3x+C

Note: The final answer in indefinite integration may vary based on the integration constant.

Page No 19.134:

Question 56:

x cos3 x dx

Answer:

Let I=x cos3x dx

As we know ,cos 3x=4 cos3 x-3cosxcos3 x=14cos 3x+3cos x

I=14x.cos 3x+3 cos xdx     =14 xI.cosII 3x dx+34  xI.cos xII dx      =14x.cos 3x dx-ddxx.cos 3x dxdx+34xcos x-ddxx.cos x dxdx     =14x.sin 3x3-1.sin 3x3dx+34xsin x-1.sin x dx     =x sin 3x12+cos 3x36+34x sin x+34cos x+C

Page No 19.134:

Question 57:

tan-1 1-x1+x dx

Answer:

Let I= tan-1 1-x1+x dxPutting x=cos θdx=-sin θ dθ& θ=cos-1 x I= tan-1 1-cos θ1+cos θ -sin θ dθ      = tan-1 2 sin2 θ22 cos2 θ2 -sin θ dθ      = tan-1 tan θ2 -sin θ dθ      =-12 θI sin θII dθ     =-12θ sinθ dθ-ddθθsin θ dθdθ      =-12 θ-cos θ- 1.-cos θ dθ      =-12 -θ cos θ+sin θ+C      =-12 -θ.cos θ+1-cos2 θ+C      =-12-cos-1 x.x+1-x2+C                 θ=cos-1 x      =x cos-1 x2-1-x22+C

Page No 19.134:

Question 58:

sin-1 xa+x dx

Answer:

Let I= sin-1 xa+x dxPutting x=a tan2 θxa=tan θdx=a2 tan θsec2 θ dθ I= sin-1 a tan2 θa+a tan2 θ 2a tan θsec2 θ dθ      = sin-1 tan2 θsec2 θ 2a tan θ sec2 θ dθ      =2a  sin-1 sin θtan θ sec2 θ dθ   

    =2a  θ Itan θ sec2 θII dθ       =2a θtan2θ2-1tan2 θ2dθ=2a θ.tan2 θ2-12sec2 θ-1dθ=a θ tan2 θ-a tan θ+aθ+C=axa tan-1 xa-axa+a tan-1 xa+C=x tan-1 xa-ax+a tan-1 xa+C

Page No 19.134:

Question 59:

x3 sin-1 x21-x4 dx

Answer:

Let I= x3 × sin-1 x21-x4dxPutting sin-1 x2=t x2=sin t 1×2x dx1-x22=dtx dx1-x4=dt2I= x2. sin-1 x21-x4.x dx     = sin t.t.dt2     =12 tI.sin tII dt     =12tsin t dt-ddttsin t dtdt     =12 t.-cos t- 1.-cos t dt     =12-t cos t+sin t+C     =12 -t1-sin2 t+sin t+C     =12 -sin-1 x2 1-x4+x2+C

Page No 19.134:

Question 60:

x2 sin-1 x1-x23/2 dx

Answer:

Let  I= x2.sin-1 x dx1-x232Putting x=sin θ dx=cos θ dθ& θ=sin-1 xI= sin2 θ.θ.cos θ dθ1-sin2 θ32    = sin2 θ.θ.cos θ dθcos2 θ32    = sin2 θ.θ.cos θ dθcos3 θ    = tan2 θ.θ.dθ    = sec2 θ-1θ.dθ    = θI.sec2 θII dθ- θ.dθ    =θsec2 θ dθ-ddθθsec2 θ dθdθ- θ.dθ    =θ tan θ- 1.tan θ dθ-θ22    =θ.tan θ-ln  sec θ-θ22+C    =θ.sin θcos θ+ln cos θ-θ22+C    =θ.sin θcos θ+ln 1-sin2 θ-θ22+C    =θ. sin θ1-sin2 θ+12ln 1-sin2 θ-θ22+C    =x sin-1 x1-x2+12ln 1-x2-12sin-1 x2+C             θ=sin-1 x



Page No 19.14:

Question 1:

3xx+4x+5dx

Answer:

    3xx+4x+5dx=3x1·x12+4x12+5dx=3x32dx+4x12dx+5dx=3x32+132+1+4x12+112+1+5x+C=3×25x52+4×23x32+5x+C=65x52+83x32+5x+C

Page No 19.14:

Question 2:

2x+5x-1x1/3dx

Answer:

    2x+5x-1x13dx=2xdx+5 dxx-dxx13=2xdx+5 dxx-x-13dx=2xln 2+5 ln x-x-13+1-13+1+C=2xln 2+5 ln x-32x23+C

Page No 19.14:

Question 3:

xax2+bx+c dx

Answer:

    x ax2+bx+cdx=x12ax2+bx+cdx=ax2+12+bx12+1+c x12dx=ax52 dx+bx32dx+cx12dx=ax52+152+1+bx32+132+1+cx12+112+1+C=2a7x72+2b5x32+2c3x32+C

Page No 19.14:

Question 4:

2-3x 3+2x 1-2x dx

Answer:

    2-3x 3+2x1-2xdx=2-3x 3-6x+2x-4x2dx=2-3x -4x2-4x+3dx=-8x2-8x+6+12x3+12x2-9xdx=12x3+4x2-17x+6dx=12x44+4x33-17x22+6x+C=3x4+43x3-172x2+6x+C

Page No 19.14:

Question 5:

mx+xm+mx+xm+mx dx

Answer:

    mx+xm+mx+xm+mxdx=m1xdx+1mxdx+mxdx+xmdx+mxdx=mlnx+1mx1+11+1+mxln m+ xm+1m+1+mx1+11+1=m ln x+x22m+mxln m+ xm+1m+1+mx22+C

Page No 19.14:

Question 6:

x-1x2 dx

Answer:

    x-1x2dx=x+1x-2dx=xdx+dxx-2dx=x22+ln x-2x+C

Page No 19.14:

Question 7:

1+x3x dx

Answer:

    1+x3xdx=1+x3+3 12x+31x2xdx=1+x3+3x+3x2x dx=1x+x3x+3xx+3x2xdx=x-12+x52+3x12+3x32dx=x-12+1-12+1+x52+152+1+3x12+112+1+3x32+132+1+C=2x+27x72+2x32+65x52+C

Page No 19.14:

Question 8:

x2+elog x+e2x dx

Answer:

    x2+elog x+e2xdx=x2dx+xdx+e2xdx=x33+x22+e2xln e2+C

Page No 19.14:

Question 9:

xe+ex+ee dx

Answer:

xe+ex+eedx=xedx+exdx+ee1dx=xe+1e+1+ex+x·ee+C

Page No 19.14:

Question 10:

xx3-2x dx

Answer:

    x x3-2xdx=x72-2xdx=x72-2x-12 dx=x72+172+1-2x-12+1-12+1+C=29x92-4x12+C=29x92-4x+C

Page No 19.14:

Question 11:

1x1+1x dx

Answer:

    1x1+1xdx=x-121+1xdx=x-12+1x32dx=x-12dx+x-32dx=x-12+1-12+1+x-32+1-32+1=2x-2x+C

Page No 19.14:

Question 12:

x6+1x2+1 dx

Answer:

     x6+1x2+1dx= x23+13x2+1dx                A3+B3=A+B A2-AB+B2=x2+1x4-x2+1x2+1dx=x4-x2+1dx=x4dx+x2dx+1dx=x4+14+1-x2+12+1+x+C=x55-x33+x+C

Page No 19.14:

Question 13:

x-1/3+x+2x3 dx

Answer:

     x-13+x+2x13dx= x-13x13+x12x13+2x13dx=x-23+x16+2x-13dx=x-23+1-23+1+x16+116+1+2x-13+1-13+1=x1313+x7676+3x23+C=3x13+67x76+3x23+C

Page No 19.14:

Question 14:

1+x2x dx

Answer:

     1+x2xdx= 1x+xx+2xxdx=x-12+x12+2dx=x-12+1-12+1+x12+112+1+2x+C=2x+23x32+2x+C



Page No 19.143:

Question 1:

ex cos x-sin x dx

Answer:

Let I=ex cos x- sin x dx let ex cos x=t Diff both sides w.r.t xex·cos x+ex-sin x=dtdx     Put ex fx=tex cos x-sin x dx=dtex cos x-sin x dx=dtI=t+C=ex cos x+C

Page No 19.143:

Question 2:

ex 1x2-2x3 dx

Answer:

Let I=ex 1x2- 2x3 dxAlso let ex×1x2=t Diff both sides w.r.t xex×1x2+ex -2x3=dtdxex 1x2-2x3 dx=dtex 1x2-2x3 dx=dt=t+C=exx2+C

Page No 19.143:

Question 3:

ex 1+sin x1+cos x dx

Answer:

Let I=ex 1+sin x1+cos x dx=ex 11+cosx +sin x1+cos x dx=ex 12 cos2 x2+2 sin x2 cos x22 cos2 x2 dx=ex 12 sec2 x2+tan x2 dx Putting ex tan x2=tDiff both sides w.r.t. xex·tan x2+ex×12sec2 x2=dtdxex tan x2+12 sec2 x2 dx=dtex 12 sec2 x2+tan x2 dx=dt=t+C=ex tanx2+C

Page No 19.143:

Question 4:

ex cot x-cosec2 x dx

Answer:

Let I=excotx-cosec2xdxhere f(x)=cotx                           put exf(x)=t         f'(x)=-cosec2xlet excotx=tDiff both sides w.r.t xexcotx+ex-cosec2x=dtdxexcotx-cosec2xdx=dtexcotx-cosec2xdx=dt=t+C=excot x+C

Page No 19.143:

Question 5:

ex x-12x2 dx

Answer:

Let I=exx-12x2dx=12ex1x-1x2dxhere 1x=f(x)                                      Put exf(x)=t-1x2=f'(x)let ex1x=tDiff both sides w.r.t xex1x+ex-1x2=dtdxex1x-1x2dx=dtI=12dt=t2+C=ex2x+C

Page No 19.143:

Question 6:

ex sec x 1+tan x dx

Answer:

Let I=exsecx1+tanxdx=exsecx+secx tanxdxHere, f(x)=secx                                    Put exf(x)=tf'(x)=secx tanxlet exsecx=tDiff both sides w.r.t xexsecx+exsecx tanx=dtdxexsecx+tanxdx=dtexsecx+secx tanxdx=dt=t+C=exsecx+C

Page No 19.143:

Question 7:

ex tan x-log cos x dx

Answer:

Let I=extanx-log cosxdxhere f(x)=-log cosx                               Put exf(x)=tf'(x)=tanxlet -exlog cosx=tDiff both sides w.r.t x-exlogcosx+ex1cosx×-sinx=dtdx-exlogcosx+extanxdx=dtextanx-log cosxdx=dt=t+C=-exlogcosx+C=exlogsec x+C

Page No 19.143:

Question 8:

ex sec x+ log sec x+tan x  dx

Answer:

Let I=exsecx+logsecx+tanxdxHere, f(x)=logsecx+tanx                                        Put exf(x)=tf'(x)=secx let exlogsecx+tanx=tDiff both sides w.r.t xexlogsecx+tanx+ex1secx+tanxsecx+tanx+sec2x=dtdxexlogsecx+tanx+exsecxdx=dtexsecx+logsecx+tanxdx=dtexsecx+logsecx+tanxdx=dt=t+C=exlogsecx+tanx+C

Page No 19.143:

Question 9:

ex cot x+log sin x dx

Answer:

Let I=excotx+log sinxdxHere,f(x)=log sinx                      Put exf(x)=tf'(x)=cotxlet exlog sinx=tDiff both sides w.r.t xexlog sinx+ex×1sinx×cosx=dtdxexlogsinx+excotxdx=dtexcotx+log sinxdx=dtexcotx+log sinxdx=dt=t+C=exlog sinx+C

Page No 19.143:

Question 10:

exx-1x+13 dx

Answer:

Let I=exx-1x-13dx=exx+1-2x+13dx=ex1x-12-2x+13dxHere, f(x)=1x+12f'(x)=-2x+12Put exf(x)=tlet ex1x+12=tDiff both sidesex1x+12+ex-2x+13=dtdxex1x+12-2x+13dx=dtex1x+12-2x+13dx=dt=t+C=exx+12+C

Page No 19.143:

Question 11:

ex sin 4x-41-cos 4x dx

Answer:

Let I=exsin4x-41-cos4xdx=ex2sin2x cos2x2sin2(2x)-42sin22xdx=excot(2x)-2cosec2(2x)dxHere, f(x)=cot(2x)f'(x)=-2cosec2(2x)Put exf(x)=tlet excot(2x)=tDiff both sides w.r.t xexcot(2x)+ex×-2cosec2(2x)=dtdxexcot(2x)-2cosec2(2x)dx=dtexcot2x-2cosec22xdx=dtI=t+C=excot2x+C

Page No 19.143:

Question 12:

ex1-x21+x22 dx

Answer:

Let I=ex1-x21+x22dx=ex1+x2-2x1+x22dx=ex11+x2-2x1+x22dxHere, f(x)=11+x2f'(x)=-2x1+x22Put exf(x)=tex11+x2=tDiff both sides w.r.t xex11+x2+ex-11+x222x=dtdxex11+x2-2x1+x22dx=dtex11+x2-2x1+x22dx=dtI=t+C=ex1+x2+C

Page No 19.143:

Question 13:

ex1+x2+x2 dx

Answer:

Let I=ex1+x2+x2dx=ex2+x-12+x2dx=ex12+x-12+x2dxHere, f(x)=12+xf'(x)=-12+x2Put exf(x)=tex1x+2=tDiff both sides w.r.t xex1x+2+ex-1x+22=dtdxex1x+2-1x+22dx=dtex12+x-12+x2dx=dtI=t+C=ex2+x+C

Page No 19.143:

Question 14:

1-sin x1+cos xe-x/2 dx

Answer:

Let I=1-sinx1+cosxe-x2dx=cos2x2+sin2x2-2sinx2cosx22cos2x2e-x2dx=cosx2-sinx222cos2x2e-x2dx=sinx2-cosx22cos2x2e-x2dx=12secx2tanx2-12secx2e-x2dx=12secx2tanx2-secx2e-x2dxlet e-x2 secx2=tDiff both sides w.r.t xe-x2secx2tanx22+secx2×e-x2×-12=dtdxe2-x2secx2tanx2-secx2dx=dt12secx2tanx2-secx2e-x2dx=dtI=t+C=e-x2secx2+C

Page No 19.143:

Question 15:

ex log x+1x dx

Answer:

Let I=exlogx+1xdxHere, f(x)=logxf'(x)=1xput exf(x)=t ex logx=tDiff both sides w.r.t xexlogx+ex1x=dtdxexlogx+1xdx=dtexlogx+1xdx=dtI=t+C=exlogx+C

Page No 19.143:

Question 16:

ex log x+1x2 dx

Answer:

Let I=logx+1x2exdx=exlogx+1x-1x+1x2dx=exlogx+1xdx+ex-1x+1x2dxlet exlogx=tDiff both sidesexlogx+ex1xdx=dtlet ex-1x=pDiff both sidesex-1x+ex1x2dx=dpI=dt+dp=t+p+C=exlogx+ex-1x+C=exlogx-1x+C

Page No 19.143:

Question 17:

exxx log x2+2 log x dx

Answer:

Let I=exxxlogx2+2logxdx=exlogx2+2logxxdxHere, f(x)=logx2f'(x)=2logxxput exf(x)=texlogx2=tDiff both sides w.r.t xexlogx2+ex2logxxdx=dtI=dt=t+C=exlogx2+C

Page No 19.143:

Question 18:

ex·1-x2 sin-1 x+11-x2 dx

Answer:

Let I=ex1-x2sin-1x+11-x2dx=exsin-1x+11-x2dxHere, f(x)=sin-1xf'(x)=11-x2Put exf(x)=t exsin-1x=tDiff both sides w.r.t xexsin-1x+ex×11-x2dx=dtI=dt=t+C=exsin-1x+C

Page No 19.143:

Question 19:

e2x -sin x+2 cos x dx

Answer:

Let I=e2x-sinx+2cosxdxPut e2xcosx=tDiff both sides w.r.t x2e2xcosx+e2x×-sinxdx=dtI=dt=t+C=e2xcosx+C

Page No 19.143:

Question 20:

1log x-1log x2 dx

Answer:

Let I=1logx-1logx2dxPut logx=tx=etdx=etdtI=et1t-1t2dtHere, f(t)=1t f'(t)=-1t2let et×1t=pDiff both sides w.r.t tet×1t+et×-1t2dt=dpI=dp=p+C=ett+C=xlogx+C

Page No 19.143:

Question 21:

ex sin x cos x-1sin2 x dx

Answer:

Let I=exsinx cosx-1sin2xdx=excotx-cosec2xdxHere, f(x)=cotxf'(x)=-cosec2xPut exf(x)=texcotx=tDiff both sides w.r.t xexcotx-cosec2xdx=dtI=dt=t+C=excot x+C

Page No 19.143:

Question 22:

tan log x+sec2 log x dx

Answer:

Let I=tanlogx+sec2logxdxPut logx=tx=etdx=etdtI=tant+sec2tetdtHere, f(t)=tantf'(t)=sec2tlet ettan(t)=pDiff both sides w.r.t tettant+sec2t=dpdtettant+sec2tdt=dpI=dp=p+C=ettant+C=x tan(logx)+C

Page No 19.143:

Question 23:

ex x-4x-23 dx

Answer:

Let I=exx-4x-23dx=exx-2-2x-23dx=ex1x-22-2x-23dxHere, f(x)=1x-22f'(x)=-2x-23Put exf(x)=t ex1x-22=tDiff both sides w.r.t xex1x-22+ex-2x-23dx=dtI=dt=t+C=exx-22+C

Page No 19.143:

Question 24:

Evaluate the following integrals:

e2x1-sin2x1-cos2xdx

Answer:

We have,I=e2x1-sin2x1-cos2xdx =e2x1-2 sinx cosx2sin2xdxPut t=2x. Then dt=2dxTherefore,I=12et1-2 sint2 cost22sin2t2dt  =14et1-2 sint2 cost2sin2t2dt  =14et1sin2t2-2 sint2cost2sin2t2dt  =14etcosec2t2-2cott2dt  =-14et2cott2-cosec2t2dtConsider, fx=2cott2, then f'x=-cosec2t2 Thus, the given integrand is of the form exfx+f'x.Therefore, I=-142cott2et+c                 =-142cot2x2e2x+cHence, e2x1-sin2x1-cos2xdx=-12cotxe2x+c



Page No 19.149:

Question 1:

eax cos bx dx

Answer:

Let I= eaxcosbxdxConsidering cos bx as first function and eax as second functionI=cosbxeaxa--sinbx×b×eaxadxI=eaxcos bxa+basin bxeax dxI=eaxacosbx+baeax×sin bxdxI=eaxacosbx+baI1                 .....1where I1=eaxsinbxdxNow, I1=eaxsin bxdxConsidering sin bx as first function eax as second functionI1=sin bxeaxa-cos bxbeaxadxI1=sin bxeaxa-baeaxcos bxdxI1=eaxsin bxa-baI                .....2From 1 and  2I=eaxacos bx+baeaxsin bxa-baII=eaxcos bxa+beaxsin bxa2-b2a2II1+b2a2=eaxa cos bx+b sin bxa2I=eaxa cos bx+b sin bxa2+b2+C

Page No 19.149:

Question 2:

eax sin bx+C dx

Answer:

Let I= eaxsin bx+CdxConsidering sin bx+C as first function and eax as second functionI=sin bx+Ceaxa- cos bx+CbeaxadxI=eaxsin bx+Ca-ba eax cos bx+C dxI=eaxsin bx+Ca-baI1              ...1where I1= eaxcos bx+CdxNow, I1= eaxcos bx+CdxConsider cos bx+C as first function eax as second funcitonI1=cos bx+Ceaxa- -sin bx+CbeaxadxI1=eaxcos bx+Ca+ba eaxsin bx+CdxI1=eaxcos bx+Ca+baI             .....2From 1 & 2I=eaxsin bx+Ca-baeaxcos bx+Ca+baII=eaxsin bx+Ca-ba2eaxcos bx+C-b2a2II1+b2a2=eax asin bx+C-beax cos bx+Ca2+C1I=eax a sin bx+C-b cos bx+Ca2+b2+C1Where C1 is integration constant

Page No 19.149:

Question 3:

cos log x dx

Answer:

Let I= cos log xdxLet log x=tx=etdx=et dtI= etcostdtConsidering cost as first function and et as second functionI=cos t et- -sin tet dtI=cos t et+ sin t et dtI=cos t et+I1             .....1where I1=etsin t dtI1= etsin t dtCosidering sin t as first function and et as second functionI1=sin t et- cos t etdtI1=sin t et-I              .....2From 1 & 2I=cos t et+sin t et-I2I=etsin t+cos tI=etsin t+cos t2+CI=elog x sinlog x+coslog x2+CI=x2sin log x+cos log x+C

Page No 19.149:

Question 4:

e2x cos 3x+4 dx

Answer:

Let I= e2xcos 3x+4dxConsidering cos 3x+4 as first function and e2x as second functionI=cos 3x+4e2x2--sin 3x+4×3e2x2dxI=e2xcos 3x+42+32 e2xsin 3x+4dxI=e2xcos 3x+42+32I1         .....1where I1= e2xsin 3x+4dxconsidering sin 3x+4 as first function and e2x as second functionI1=sin 3x+4e2x2- 3 cos 3x+4e2x2dxI1=e2x sin 3x+42-32 e2xcos 3x+4dxI1=e2xsin 3x+42-32 I           .....2From 1 & 2I=e2xcos 3x+42+34 e2xsin 3x+4-94II+94I=2e2xcos3x+4+3e2xsin3x+44I=e2x132 cos 3x+4+3 sin 3x+4+C

Page No 19.149:

Question 5:

e2x sin x cos x dx

Answer:

Let I= e2xsin x cos x dxI=12 e2x 2 sin x cos xdxI=12 e2xsin2xdxConsidering sin2x as first function and e2x as second function.I=12sin2xe2x2-2cos2xe2x2dxI=e2xsin2x4-12 e2xcos2xdxI=e2xsin2x4-12I1          .....1Where I1= e2xcos2xdxConsidering cos2x as first function and e2x as second functionI1=cos2xe2x2--2 sin2xe2x2dxI1=e2xcos2x2+ e2xsin2x dxI1=e2xcos2x4+2I         .....2I=e2xsin2x4-12e2xcos2x2+2II=e2xsin2x4-e2xcos2x4-I2×22I=e2x sin2x-cos2x4+CI=e2x8sin2x-cos2x+C

Page No 19.149:

Question 6:

e2x sin x dx

Answer:

Let I= e2xsin x dxConsidering sin x as first function and e2x as second functionI=sin xe2x2- cos xe2x2dxI=sin xe2x2-12 cos x e2x dxI=sin x e2x2-12cos xe2x2--sin xe2x2dxI=sin x e2x2-cos x e2x4-12e2xsin x2dxI=e2x2 sin x-cos x4-I45I=e2x 2 sin x-cos xI=e2x 2 sin x-cos x5+C

Page No 19.149:

Question 7:

Evaluate the following integrals:

e2x sin3x+1 dx

Answer:

We have,I=e2x sin3x+1 dxLet the first function be sin3x+1 and the second function be e2x.First we find the integral of the second function, i.e., e2x dx.e2x dx=12e2xNow, using integration by parts, we getI=sin3x+1e2x dx-dsin3x+1dxe2x dxdx =12sin3x+1e2x-32cos3x+1 e2xdx =12sin3x+1e2x-32cos3x+1e2x dx-dcos3x+1dxe2x dxdx =12sin3x+1e2x-3212cos3x+1e2x+32sin3x+1 e2x dx =12sin3x+1e2x-34cos3x+1e2x-94I+cI+94I=12sin3x+1e2x-34cos3x+1e2x+c134I=e2x2sin3x+1-32cos3x+1+cI=213e2xsin3x+1-32cos3x+1+c =e2x132 sin3x+1-3 cos3x+1+cHence, e2x sin3x+1 dx=e2x132 sin3x+1-3 cos3x+1+c

Page No 19.149:

Question 8:

ex sin2 x dx

Answer:

Let I= exsin2 x dx= ex 1-cos 2x2dx=12 ex dx-12 ex cos 2x dx=ex2-12 excos 2x dx         .....1Let I1= excos 2xdxConsidering cos 2x as first function and ex as second functionI1=cos 2xex--2 sin 2xex dxI1=cos 2xex+2 sin 2xex dxI1=cos 2xex+2sin 2xex- 2 cos 2xex dxI1=cos 2xex+2 sin 2xex-4I15I1=ex cos 2x+2 sin 2xI1=ex5cos 2x+2 sin 2x+C       .....2From 1 & 2I=ex2-ex10cos 2x+2 sin 2x+C

Page No 19.149:

Question 9:

1x3sin log x dx

Answer:

Let I= 1x3sin log xdxPutting log x=tx=etdx=et dtI= 1e3tsin t et dt=e-2t sin t dtConsidering sin t as first function and e-2t as second functionI=sin te-2t-2- cos te-2t-2dtI=sin t e-2t-2+12cos t e-2t dtI=sin t e-2t-2+12cos te-2t-2--sin te-2t-2dtI=sin t e-2t-2-14 cos t e-2t- e-2tsin t dt4I=e-2t -2 sin t-cos t4-I45I4=e-2t -2 sin t-cos t4I=e-2t5-2 sin t-cos t+CI=-x-252 sin log x+cos log x+CI=-15x2cos log x+2 sin log x+C

Page No 19.149:

Question 10:

e2x cos2 x dx

Answer:

Let I= e2xcos2 x dx= e2x1+cos 2x2dx=12 e2xdx +12e2xcos 2xdx=e2x4+12I1         .....1Where I1= e2xcos 2x dxConsidering cos 2x as first function and e2x as second function I1=cos 2xe2x2--2 sin 2x×e2x2dx I1=cos 2xe2x2+ e2xsin 2x dxNow considering sin 2x as first function and e2x as second function I1=cos 2xe2x2+sin 2xe2x2- 2 cos 2xe2x2dx I1=e2xcos 2x+sin 2x2-I12 I1=e2xcos 2x+sin 2x2 I1=e2xcos 2x+sin 2x4          .....2From 1 &  2I=e2x4+e2x8cos 2x+sin 2x+C

Page No 19.149:

Question 11:

e-2x sin x dx

Answer:

Let I= e-2xsin x dxConsidering sin x as first function and e-2x as second functionI=sin xe-2x-2-cos xe-2x-2dxI=-e-2xsin x2+12e-2x cos x dxI=-e-2xsin x2+I12 where           .....1Where, I1= e-2xcos x dxConsidering cos x as first and e-2x as second functionI1=cos x e-2x-2--sin xe-2x-2dxI1=e-2xcos x-2-sin x e-2x dx2I1=-e-2x cos x2-I2           .....2From 1 & 2I=-e-2x sin x2+12-e-2xcos x2-I2I+I4=-e-2xsin x2-e-2xcos x45I4=-e-2x 2 sin x+cos x4I=e-2x5-2 sin x-cos x+C

Page No 19.149:

Question 12:

x2 ex3 cos x3 dx

Answer:

Given integral is, x2ex3cos x3 dxLet x3=t3x2 dx=dtx2 dx=dt3Integral becomes,13 etcos t dt=13I                 .....1Where, I= etcos t dtI= etcos t dtConsidering cos t as first and et as second functionI=cos t et--sin t et dtI=etcos t+ sin t et dtAgain considering sin t as first and et as second functionI=etcos t+sin t et- cos t et dtI=et cos t+sin t et-I2I=etsin t+cos tI=et2sin t+cos t x2ex3cos x3 dx=13et2sin t+cos t+C                  From 1=ex36sin x3+cos x3+C



Page No 19.15:

Question 15:

x3-5x dx

Answer:

    x3-5xdx= x123-5xdx=3x12-5x32dx=3x12+112+1-5x32+132+1+C=2x32-2x52+C

Page No 19.15:

Question 16:

x+1x-2x dx

Answer:

    x+1x-2xdx= x2-2x+x-2xdx=x2-x-2xdx=x32-x12-2x-12dx=x32+132+1-x12+112+1-2x-12+1-12+1+C=25x52-23x32-4x12+C

Page No 19.15:

Question 17:

x5+x-2+2x2 dx

Answer:

    x5+x-2+2x2dx= x5x2+x-2x2+2x2dx=x3+x-4+2x-2dx=x3+13+1+x-4+1-4+1+2x-2+1-2+1+C=x44-13x3-2x+C

Page No 19.15:

Question 18:

3x+42 dx

Answer:

    3x+42dx= 9x2+2×3x×4+16dx=9x2dx+24x dx+16dx=9x33+24x22+16x+C=3x3+12x2+16x+C

Page No 19.15:

Question 19:

2x4+7x3+6x2x2+2x dx

Answer:

    2x4+7x3+6x2x2+2xdx= x22x2+7x+6xx+2dx=x2x2+4x+3x+6x+2dx=x2xx+2+3x+2x+2dx=x2x+3x+2x+2dx=2x2+3xdx=2x2dx+3x dx=2x33+3x22+C=23x3+32x2+C

Page No 19.15:

Question 20:

5x4+12x3+7x2x2+x dx

Answer:

    5x4+12x3+7x2x2+xdx= x25x2+12x+7xx+1dx=x5x2+5x+7x+7x+1dx=x5xx+1+7x+1x+1dx=x5x+7x+1x+1dx=5x2+7xdx=5x33+7x22+C

Page No 19.15:

Question 21:

sin2 x1+cos x dx

Answer:

      sin2x1+cos xdx=1-cos2x1+cos xdx=1-cos x 1+cos x1+ cos xdx= 1-cos xdx=x-sin x+C

Page No 19.15:

Question 22:

sec2 x+cosec2 x dx

Answer:

    sec2x+cosec2xdx=sec2x dx+cosec2x dx=tan x-cot x+C

Page No 19.15:

Question 23:

sin3 x-cos3 xsin2x cos2 x dx

Answer:

    sin3 x-cos3 xsin2 x · cos2 xdx=sin3xsin2 x · cos2xdx-cos3 xsin2 x · cos2 xdx=sin xcos2 xdx-cos xsin2 xdx=sin xcos x×1cos xdx-cos xsin x×1sin xdx=sec x tan x dx-cosec x cot x dx=sec x--cosec x+C=sec x+cosec x+C

Page No 19.15:

Question 24:

5 cos3 x+6 sin3 x2 sin2 x cos2 x dx

Answer:

    5 cos3x+6 sin3x2 sin2x cos2xdx=5 cos3x2 sin2x cos2x+6 sin3x2sin2x cos2xdx=52 cos xsin2x+3sin xcos2xdx=52cos xsin x×1sin xdx+3sin xcos x×1cos xdx=52cosec x cot x dx+3sec x tan x dx=52-cosec x+3 sec x+C=-52cosec x+3 sec x+C

Page No 19.15:

Question 25:

tan x+cot x2 dx

Answer:

    tan x+cot x2=tan2x+cot2x+2 tan x cot xdx=tan2x+cot2x+2dx=sec2x-1+cosec2x-1+2dx=sec2x+cosec2x dx=tan x -cot x+C

Page No 19.15:

Question 26:

1-cos 2x1+cos 2x dx

Answer:

    1-cos 2x1+cos 2xdx=2 sin2x2 cos2xdx                     1-cos 2θ=2 sin2 θ   & 1+cos 2θ=2 cos2θ=tan2x dx           =sec2x-1 dx=sec2x dx-dx=tan x -x+C

Page No 19.15:

Question 27:

cos x1-cos x dx or cot xcosec x-cot x dx

Answer:

    cot xcosec x-cot xdx=cos xsin x1sin x-cos xsin xdx=cos x1-cos x×1+cos x1+cos xdx=cos x+cos2x1-cos2xdx=cos x+cos2xsin2x dx=cos xsin x×1sin x+cos2xsin2xdx=cot x cosec x+cot2xdx=cosec x cot x+cosec2x-1dx=-cosec x-cot x-x+C

Page No 19.15:

Question 28:

cos2 x-sin2 x1+cos 4x dx

Answer:

     cos2x-sin2x1+cos 4xdx=cos 2x2 cos2 2xdx     1+cos A=2 cos2 A2  &  cos2A-sin2A=cos2A=12cos 2xcos 2xdx=12x+C=x2+C

Page No 19.15:

Question 29:

11-cos x dx

Answer:

      dx1-cos x=dx1-cos x×1+cosx1+cosx=1+cos x1-cos2xdx=1+cos xsin2xdx=1sin2x+cos xsin x×1sin xdx=cosec2x+cosec x cot xdx=-cot x-cosec x+C

Page No 19.15:

Question 30:

11-sin x dx

Answer:

      dx1-sin x=1+sin x1-sin x×1+sin xdx=1+sin x1-sin2xdx=1+sin xcos2xdx=1cos2x+sin xcos x×1cos xdx=sec2x+sec x tan xdx=tan x+sec x+C

Page No 19.15:

Question 31:

tan xsec x+tan x dx

Answer:

      tan xsec x+tan xdx=tan xsec x+tan x×sec x-tan xsec x-tan xdx=tan x sec x-tan xsec2x-tan2xdx=sec x tan x-tan2x1dx=sec x tan x dx-sec2x-1dx=sec x-tan x+x+C

Page No 19.15:

Question 32:

cosec xcosec x-cot x dx

Answer:

      cosec xcosec x-cot xdx=cosec xcosec x+cot xcosec x-cot x cosec x+cot xdx=cosec x cosec x+cot xcosec2x-cot2xdx=cosec2x+cosec x cot xdx=-cot x-cosec x+C

Page No 19.15:

Question 33:

11+cos 2x dx

Answer:

      dx1+cos 2x               1+cosθ=2cos2 θ2=dx2 cos2x=12sec2x dx=12tan x+C

Page No 19.15:

Question 34:

11-cos 2x dx

Answer:

      dx1-cos 2x               1-cos A=2sin2 A2=dx2 sin2x=12cosec2x dx=12-cot x+C=-12cot x+C

Page No 19.15:

Question 35:

tan-1sin 2x1+cos 2x dx

Answer:

      tan-1sin 2x1+cos2xdx=tan-12 sin x cos x2 cos2xdx             sin 2x=2 sinx cosx   &  1+cos2x=2 cos2x=tan-1 tan x=tan-1 tan x=x dx=x22+C

Page No 19.15:

Question 36:

cos-1sin x dx

Answer:

      cos-1 sin xdx=cos-1cosπ2-xdx              sin x=cosπ2-x=π2-xdx=πx2-x22+C

Page No 19.15:

Question 37:

cot-1sin 2x1-cos 2x dx

Answer:

      cot-1 sin 2x1-cos 2xdx=cot-12 sin x cos x2 sin2xdx      sin 2x=2 sin x cos x   &   1-cos 2x=2 sin2x=cot-1 cot xdx=x dx=x22+C

Page No 19.15:

Question 38:

sin-12 tan x1+tan2 x dx

Answer:

      sin-1 2 tan x1+tan2xdx=sin-1 sin 2 xdx     sin 2x=2 tan x1+tan2x=2x dx=2 x22+C=x2+C

Page No 19.15:

Question 39:

x3+8x-1x2-2x+4 dx

Answer:

      x3+8x-1x2-2x+4dx=x3+23 x-1x2-2x+4dx           =x+2 x2-2x+4 x-1x2-2x+4 dx        a3+b3=a+b a2-ab+b2= x+2 x-1dx=x2-x+2x-2dx=x2+x-2dx=x2 dx+x dx-2 1dx=x33+x22-2x+C

Page No 19.15:

Question 40:

a tan x+b cot x2 dx

Answer:

      a tan x+b cot x2dx=a2 tan2x+b2 cot2x+2ab tan x cot xdx=a2tan2x dx +b2 cot2x dx+2ab dx=a2sec2x-1dx+b2cosec2x-1dx+2abdx=a2tan x-x +b2 -cot x-x+2ab x+C=a2 tan x -b2cot x-a2+b2-2abx+C

Page No 19.15:

Question 41:

x3-3x2+5x-7+x2ax2x2 dx

Answer:

      x3-3x2+5x-7+x2ax2x2dx=x32x2-3x22x2+5x2x2-72x2+x2ax2x2dx=x2-32+52x-72x-2+ax2dx=12x dx-32dx+52dxx-72x-2 dx+12axdx=12x22-32x+52ln x-72 x-2+1-2+1+12axln a+C=x24-32x+52ln x+72x+ax2 ln a+C=12x22-3x+5 ln x+7x+ax ln a+C

Page No 19.15:

Question 42:

cos x1+cos x dx

Answer:

      cos x1+cos xdx=cos x1-cos x1+cos x1-cos xdx=cos x-cos2x1-cos2 xdx=cos x-cos2 xsin2 xdx=cos xsin2 x-cos2 xsin2 xdx=cot x cosec x - cot2 xdx=cot x cosec x-cosec2 x+1dx=cot x cosec x dx-cosec2 x dx+1dx=-cosec x+cot x+x+C

Page No 19.15:

Question 43:

1-cos x1+cos x dx

Answer:

      1-cos x1+cos xdx=1-cos x21-cos2 xdx=1+cos2 x-2cos xsin2 xdx= 1sin2 x+cos2 xsin2 x-2cos xsin2 xdx= cosec2 x+cot2 x-2cot x.cosec xdx= cosec2 x+cosec2 x -1-2cot x.cosec xdx= 2cosec2 x-1-2cot x.cosec xdx=2cosec2 x dx-1 dx-2cot x.cosec x dx=-2cot x-x+2cosec x+C=2cosec x-cot x-x+C

Page No 19.15:

Question 45:

If f' (x) = x1x2 and f(1) = 12, find f(x)

Answer:

     f'x=x-1x2     f'x=x-x-2 f'xdx =x-x-2dx fx=x22-x-2+1-2+1+C       =x22+1x+Cf1=12      Given122+11+C=12C=-1 fx=x22+1x-1

Page No 19.15:

Question 46:

If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)

Answer:

     f'x=x+b, f1=5, f2=13    f'x=x+bf'xdx =x+bdxfx=x22+bx+C     ....(i)f1=5, f2=13       GivenPuting x=1 in (i)f1=122+b1+C5=12+b+C     ...   iiPuting x=2 in (i)f2=222+b2+C13=42+2b+C13=2+2b+C       ...(iii)Solving (ii) and (iii) we get,b=132 and C=-2Thus, fx=x22+132x-2

Page No 19.15:

Question 47:

If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)

Answer:

f'x=8x3-2x  f2=8f'x=8x3-2xf'xdx =8x3-2xdx                  =8x3dx-2xdxfx=8 x44-2×x22+Cfx=2x4-x2+Cf2=8                  Givenf2=2×24-22+C8=32-4+CC=-20fx=2x4-x2-20

Page No 19.15:

Question 48:

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, fπ2 = 5, find f(x)

Answer:

f'x=a sinx+b cos xf'0=4, f0=3fπ2=5f'x=a sin x+b cos xf'xdx=a sin x+b cos xdxfx=-a cos x+b sin x +C      ...(i)Now puting x=0 in equation (i)f0=-a cos 0+b sin 0 +C3=-a×1+b×0 + C3=-a+C    ...  iiNow puting x=π2 in equation (i)fπ2=-a cos π2+b sin π2 +C5=-a cosπ2+b sin π2+C5=-a×0+b×1+C5=b+C    ...   iiiWe also have f'0=4f'x=a sin x+b cos xf'0=a sin 0+b cos 04=a×0+b×14=b     ...  ivsolving ii, iii  and iv we get,b=4C=1a =-2 fx=2cos x+4 sin x +1

Page No 19.15:

Question 49:

Write the primitive or anti-derivative of fx=x+1x.

Answer:

fx=x+1x
Integrating both sides
fxdx=x+1xdx=x12+x-12dx=x12+112+1+x-12+1-12+1+C=23x32+2x12+C



Page No 19.154:

Question 1:

3+2x-x2 dx

Answer:

 3+2x-x2 dx= 3-x2-2x dx= 3-x2-2x+1-1 dx= 4-x-12 dx= 22-x-12 dx                                       a2-x2dx=12xa2-x2+12a2sin-1xa+C=x-12 22-x-12+222sin-1 x-12+C=x-123+2x-x2+sin-1 x-12+C

Page No 19.154:

Question 2:

x2+x+1 dx

Answer:

 x2+x+1 dx= x2+x+122-122+1 dx= x+122+322=x+122 x+122+322+38log x+12+12+x2+x+1+C          x2+a2dx=12xx2+a2+12a2 lnx+x2+a2+C=2x+14 x2+x+1+38log 2x+1+12+x2+x+1+C

Page No 19.154:

Question 3:

x-x2 dx

Answer:

 x-x2 dx= -x2-x dx= -x2-x+122-122 dx= 122-x-122 dx=x-122 x-x2 +18sin-1 x-1212+C                a2-x2dx=12xa2-x2+12a2sin-1xa+C=2x-14 x-x2+18 sin-1 2x-1+C

Page No 19.154:

Question 4:

1+x-2x2 dx

Answer:

 1+x-2x2 dx= 212+x2-x2 dx=2 12-x2-x2 dx=2  12-x2-x2+142-142 dx=2  12+116-x-142 dx=2  342-x-142 dx=2 x-142 342-x-142+932sin-1x-1434+C          a2-x2dx=12xa2-x2+12a2sin-1xa+C=4x-18 1+x-2x2+9232sin-1 4x-13+C

Page No 19.154:

Question 5:

cos x 4-sin2 x dx

Answer:

Let I= cos x 4-sin2 x dxPutting sin x=tcos x dx=dt 22-t2 dt=t222-t2+222 sin-1 t2+C                 a2-x2 dx=12xa2-x2+12a2sin-1xa+C=sin x2 4-sin2 x+2 sin-1 sin x2+C       t= sin x

Page No 19.154:

Question 6:

ex e2x+1 dx

Answer:

Let I= ex e2x+1 dxPutting ex=tex dx=dtI= t2+1dt     =t2t2+1+122ln t+t2+1+C         x2+a2dx=12xx2+a2+12lnx+x2+a2+C       =ex2 e2x+1+12ln ex+e2x+1+C                t= ex

Page No 19.154:

Question 7:

9-x2 dx

Answer:

 9-x2 dx= 32-x2 dx=x232-x2+322sin-1 x3+C        a2-x2 dx=12xa2-x2+12a2 sin-1xa+C=x29-x2+92sin-1 x3+C

Page No 19.154:

Question 8:

16x2+25 dx

Answer:

 16x2+25 dx= 16x2+2516dx=4 x2+542 dx=4x2x2+542+5422ln x+x2+542+C         x2+a2 dx=12xx2+a2+12a2 lnx+x2+a2+C=2x x2+2516+258ln x+x2+2516+C

Page No 19.154:

Question 9:

4x2-5 dx

Answer:

 4x2-5dx= 4x2-54 dx=2 x2-522 dx=2x2x2-54-58ln x+x2-54 +C        x2-a2 dx=12xx2-a2-12a2 lnx+x2-a2+C=x x2-54-54ln x+x2-54+C

Page No 19.154:

Question 10:

2x2+3x+4 dx

Answer:

 2x2+3x+4 dx=2  x2+32x+2 dx=2  x2+32x+342-342+2 dx=2  x+342-916+2 dx=2  x+342+2342dx=2 x+342 x+342+2342+2332ln x+34+x2+32x+2+C          x2+a2 dx=12xx2+a2-12a2 lnx+x2+a2+C=2 4x+38 x2+32x+2+2332ln x+34+x2+32x+2+C=4x+38 2x2+3x+4+23232ln x+34+x2+32x+2+C

Page No 19.154:

Question 11:

3-2x-2x2 dx

Answer:

Let I=3-2x-2x2dx=3-2x2+2xdx=3-2 x2+xdx=3-2 x2+x+14-14dx=3-2 x+122+12dx=72-2 x+122dx=274-x+122dx=2722-x+122dx=2×x+122 722-x+122+2×74×2 sin-1 x+1272+C=2x+14 3-2x-2x2+742 sin-1 2x+17+C

Page No 19.154:

Question 12:

xx4+1 dx

Answer:

Let I=xx4+1dx=xx22+1dxPutting x2=t2x dx=dtx dx=dt2I=12t2+1dt=12t2+12dt=12 t2t2+1+122 log t+t2+1+C=12x22 x4+1+12 log x2+x4+1+C=x24 x4+1+14 log x2+x4+1+C

Page No 19.154:

Question 13:

x2a6-x6 dx

Answer:

Let I=x2a6-x6dx=x2a32-x32dxPutting x3=t3x2 dx=dtx2 dx=dt3I=13a32-t2dt=13 t2a32-t2+a322 sin-1 ta3+C=x36 a6-x6+a66 sin-1 x3a3+C

Page No 19.154:

Question 14:

16+log x2x dx

Answer:

Let I=16+log x2xdxPutting log x=t1x dx=dtI=16+t2dt=42+t2dt=t2 42+t2+422 log t+42+t2+C=log x2 16+log x2+8 log log x+16+log x2+C



Page No 19.155:

Question 15:

2ax-x2 dx

Answer:

Let I=2ax-x2dx=a2+2ax-x2-a2dx= a2-x2-2ax+a2dx=a2-x-a2dx=x-a2 2ax-x2+a22 sin-1 x-aa+C

Page No 19.155:

Question 16:

3-x2 dx

Answer:

Let I= 3-x2dx=32-x2dx=x232-x2+322 sin-1 x3+C=x2 3-x2+32 sin-1 x3+C

Page No 19.155:

Question 17:

x2-2x dx

Answer:

I=x2-2xdx

I =x2-2x+1-1dxI=(x-1)2-12dx x2-a2dx=x2x2-a2-a22lnx+x2-a2+c I=(x-1)2(x-1)2-1-12lnx-1+x2-2x+c

Page No 19.155:

Question 18:

2x-x2 dx

Answer:

I=2x-x2dx=x2-xdx

Let x=1+sinu

or, dx=cosu duI=1+sinu1-sinucosu duI=cos2u duI=12cos2u+1du

I=1212sin 2u+u+cI=12sinu cosu+u+cI=12sinu 1-sin2u+u+c I=12x-12x-x2+12sin-1x-1+c           u=sin-1x-1



Page No 19.158:

Question 1:

x+1 x2-x+1 dx

Answer:

Let I=  x+1 x2-x+1 dxAlso, x+1=λddxx2-x+1+μx+1=λ2x-1+μx+1=λ2x-1+μx+1=2λx+μ-λEquating the coefficient of like terms2λ=1λ=12Andμ-λ=1μ-12=1μ=32I=122x-1+32 x2-x+1dx=122x-1 x2-x+1dx+32x2-x+1dx=122x-1 x2-x+1 dx+32 x2-x+14-14+1 dx=122x-1 x2-x+1 dx+32x-122+34dx=122x-1 x2-x+1 dx+32x-122+322dxLet x2-x+1=t2x-1dx=dtI=12t dt+32x-122+322 dx=12×t3232+32x-122 x-122+322+38log x-12+x-122+322+C=13x2-x+132+382x-1 x2-x+1+916log x-12+x2-x+1+C

Page No 19.158:

Question 2:

x+1 2x2+3 dx

Answer:

Let I= x+1 2x2+3 dxAlso, x+1=λddx2x2+3+μx+1=λ4x+μEquating coefficient of like terms4λ=1λ=14 and μ=1I= 144x+1 2x2+3 dx=14 4x 2x2+3 dx+2x2+3 dx=144x2x2+3dx+2x2+32dx=144x 2x2+3 dx+2 x2+322 dxLet 2x2+3=t4x dx=dtI=14 t dt+2x2x2+32+34log x+x2+32=14t3232+2x22x2+32+324 log x+2x2+32+C=162x2+332+x22x2+3+324log 2x+2x2+32+C=162x2+332+x22x2+3+324log 2x+2x2+3-324log 2+C=162x2+332+x22x2+3+324log 2x+2x2+3+C'Where C'=C-324log 2



Page No 19.159:

Question 3:

2x-5 2+3x-x2 dx

Answer:

Let I= 2x-5 2+3x-x2 dxAlso, 2x-5=λddx2+3x-x2+μ2x-5=λ-2x+3+μ2x-5=-2λx+3λ+μEquating coeffieicents of like terms-2λ=2λ=-1And3λ+μ=-53-1+μ=-5μ=-5+3μ=-22x-5=-1-2x+3-2Hence, I= --2x+3-2 2+3x-x2 dx=- -2x+3 2+3x-x2dx-22+3x-x2 dx=-I1-2I2                                                 .....1I1=-2x+3 2+3x-x2 dxLet 2+3x-x2=t-2x+3dx=dtI1= t12 dt=t12+112+1=23t32=232+3x-x232              .....2And I2= 2+3x-x2 dxI2= 2-x2-3x dx= 2-x2-3x+322-322 dx=2+94-x-322 dx= 1722-x-322 dx=x-322 1722-x-322+17222sin-1 x-32172=2x-342+3x-x2+178 sin-1 2x-317                 .....3From eq 1, 2 and 3 we haveI=-23 2+3x-x232-2x-322+3x-x2-174sin-1 2x-317+C

Page No 19.159:

Question 4:

x+2 x2+x+1 dx

Answer:

Let I= x+2 x2+x+1 dxAlso,  x+2=λddx x2+x+1+μx+2=λ2x+1+μx+2=2λx+λ+μEquating coefficient of like terms2λ=1 λ=12And λ+μ=212+μ=2μ=32I= 122x+1+32x2+x+1dx=12 2x+1 x2+x+1 dx+32x2+x+1 dx=122x+1 x2+x+1dx+32x2+x+122-122+1dx=12 2x+1 x2+x+1 dx+32 x+122+322 dxLet x2+x+1=t2x+1dx=dtThen,I=12 t dt+32 x+122+322 dx=12 t12 dt+32 x+122 x+122+322+38log x+12+ x+122+322+C=12t3232+382x+1 x2+x+1+916log x+12+x2+x+1+C=13 x2+x+132+38 2x+1 x2+x+1+916log x+12+x2+x+1+C

Page No 19.159:

Question 5:

4x+1 x2-x-2 dx

Answer:

Let I= 4x+1 x2-x-2 dxAlso, 4x+1=λddxx2-x-2+μ4x+1=λ2x-1+μ4x+1=2λx+μ-λEquating coefficient of like terms2λ=4λ=2Andμ-λ=1μ-2=1μ=3I= 22x-1+3 x2-x-2 dx=22x-1 x2-x-2dx+3x2-x-2dx=22x-1 x2-x-2 dx+3  x2-x+122-122-2 dx=2  2x+1 x2-x-2 dx+3  x-122-2-14 dx= 2x-1 x2-x-2 dx+3 x-122-322 dxLet x2-x-2=t 2x-1dx=dtI=2 t dt+3x-122 x-122-322-3222log x-12+x2-x-2=2 t3232+34 2x-1 x2-x-2-278log x-12+x2-x-2+C=43 x2-x-232+34 2x-1 x2-x-2-278log x-12+x2-x-2+C

Page No 19.159:

Question 6:

x-2 2x2-6x+5 dx

Answer:

Let I= x-2 2x2-6x+5 dxAlso, x-2=λddx2x2-6x+5+μx-2=4λx+μ-6λEquating the coefficient of like terms4λ=1λ=14Andμ-6λ=-2μ-6×14=-2μ=-2+32=-12I= 144x-6-12 2x2-6x+5 dx=14  4x-6 2x2-6x+5 dx-122x2-6x+5 dxLet 2x2-6x+5=t4x-6dx=dtI=14 t12 dt-122x2-3x+52dx=14 t12-22x2-3x+322-322+52 dx=14t3232-12 x-322-94+52 dx=16t32-12  x-322-9+104 dx=16t32-12x-322+122 dx=162x2-6x+532-12 x-322 x-322+122+18log x-32+x2-3x+52+C=16 2x2-6x+532-12 2x-34 x2-3x+52+18log 2x-32+x2-3x+52+C

Page No 19.159:

Question 7:

x+1 x2+x+1 dx

Answer:

Let I= x+1 x2+x+1 dxAlso, x+1=λddx x2+x+1+μx+1=λ2x+1+μx+1=2λx+λ+μEquating coefficient of like terms2λ=1  λ=12Andλ+μ=112+μ=1μ=12I= 12 2x+1 x2+x+1dx+12x2+x+1 dx=122x+1 x2+x+1 dx+12x2+x+122-122+1 dx=122x+1 x2+x+1 dx+12x+122+322 dxLet x2+x+1=t2x+1dx=dtThen,
I=12t dt+12x+122x+122+322+38log x+12+x+122+322+C=12×23t32+122x+14 x2+x+1+38log x+12+x2+x+1+C=13x2+x+132+122x+14 x2+x+1+38log x+12+x2+x+1+C

Page No 19.159:

Question 8:

2x+3 x2+4x+3 dx

Answer:

Let I= 2x+3 x2+4x+3 dxAlso, 2x+3=λddxx2+4x+3+μ2x+3=λ2x+4+μ2x+3=2λx+4λ+μEquating coefficient of like terms.2λ=2  λ=1And4λ+μ=34+μ=3μ=-1I= 2x+4-1 x2+4x+3dx= 2x+4 x2+4x+3dx-x2+4x+3 dx= 2x+4 x2+4x+3 dx-x2+4x+4-1 dx=2x+4 x2+4x+3dx-x+22-12 dxLet x2+4x+3=t2x+4dx=dtThen,I=t dt-x+22-12 dx=23t32-x+22x+22-1-122log x+2+x+22-1+C=23 x2+4x+332-12x+2 x2+4x+3-log x+2+ x2+4x+3+C

Page No 19.159:

Question 9:

2x-5 x2-4x+3 dx

Answer:

Let I= 2x-5 x2-4x+3 dx= 2x-4-1 x2-4x+3 dx=2x-4 x2-4x+3 dx-x2-4x+3 dx=2x-4 x2-4x+3 dx- x2-4x+4-4+3 dx=2x-4 x2-4x+3 dx- x-22-12 dxLet x2-4x+3=t2x-4dx=dtI=t dt-x-22-12 dx=23t32-x-22 x-22-12-122logx-2+x-22-1+C=23x2-4x+332-x-22 x2-4x+3+12log x-2+x2-4x+3+C

Page No 19.159:

Question 10:

xx2+x dx

Answer:

Let I= xx2+xdxAlso, x=λddxx2+x+μx=λ2x+1+μx=2λx+λ+μEquating coefficient of like terms2λ=1λ=12Andλ+μ=0μ=-12I= 122x+1-12 x2+xdx=122x+1 x2+xdx-12x2+xdx=12 2x+1 x2+xdx-12x2+x+14-14dx=122x+1 x2+x dx-12x+122-122dxLet x2+x=t2x+1dx=dtThen,I=12t dt-12x+122 x2+x-18logx+12+x2+x+C=12×23t32-2x+18 x2+x+116log x+12+x2+x+C=13x2+x32-2x+18 x2+x+116log x+12+x2+x+C

Page No 19.159:

Question 11:

Evaluate the following integrals:

x-3x2+3x-18 dx

Answer:

Let I=x-3x2+3x-18 dxWe express x-3=Addxx2+3x-18+Bx-3=A(2x+3)+BEquating the coefficients of x and constants, we get1=2A      and     -3=3A+Bor A=12      and     B=-92 I=122x+3-92x2+3x-18 dx     =122x+3x2+3x-18 dx-92x2+3x-18 dx     =12I1-92I2       ...(1)Now, I1=2x+3x2+3x-18 dx      Let x2+3x-18=u      On differentiating both sides, we get      2x+3dx=du I1=udu       =23u32+c1       =23x2+3x-1832+c1      ...(2)And, I2=x2+3x-18 dx          =x2+3x+94-94-18 dx          =x+322-922 dx      Let x+32=u      On differentiating both sides, we get      dx=du I2=u2-922 du       =u2u2-922-12922logu+u2-922+c2       =x+322x+322-922-12922logx+32+x+322-922+c2       =2x+34x2+3x-18-818logx+32+x2+3x-18+c2      ...(3)From (1), (2) and (3), we get I=1223x2+3x-1832+c1-922x+34x2+3x-18-818logx+32+x2+3x-18+c2      =13x2+3x-1832-982x+3x2+3x-18+72916logx+32+x2+3x-18+cHence, x-3x2+3x-18 dx=13x2+3x-1832-982x+3x2+3x-18+72916logx+32+x2+3x-18+c

Page No 19.159:

Question 12:

Evaluate the following integrals:

x+33-4x-x2 dx

Answer:

Let I=x+33-4x-x2 dxWe express x+3=Addx3-4x-x2+Bx+3=A(-4-2x)+BEquating the coefficients of x and constants, we get1=-2A      and     3=-4A+Bor A=-12      and     B=1 I=-12-4-2x+13-4x-x2 dx     =-12-4-2x3-4x-x2 dx+3-4x-x2 dx     =-12I1+I2       ...(1)Now, I1=-4-2x3-4x-x2 dx      Let 3-4x-x2=u      On differentiating both sides, we get      -4-2xdx=du I1=udu       =23u32+c1       =233-4x-x232+c1      ...(2)And, I2=3-4x-x2 dx          =3+4-4-4x-x2 dx          =72-x+22 dx      Let x+2=u      On differentiating both sides, we get      dx=du I2=72-u2 du       =u272-u2+1272sin-1u7+c2       =x+227-x+22+72sin-1x+27+c2       =x+223-4x-x2+72sin-1x+27+c2      ...(3)From (1), (2) and (3), we get I=-12233-4x-x232+c1+x+223-4x-x2+72sin-1x+27+c2      =-133-4x-x232+x+223-4x-x2+72sin-1x+27+cHence, x+33-4x-x2 dx=-133-4x-x232+x+223-4x-x2+72sin-1x+27+c

Page No 19.159:

Question 13:

(3x+1) 4-3x-2x2 dx

Answer:

I=3x+14-3x-2x2 dxLet 3x+1=Addx4-3x-2x2+B3x+1=A-3-4x+B3x+1=-4Ax+B-3A3=-4A and B-3A=1A=-34 and B=-54

3x+1=-34-3-4x-54I=-34-3-4x4-3x-2x2dx-544-3x-2x2dxLet I=-34I1-54I2               ...iNow,I1=-3-4x4-3x-2x2dxLet 4-3x-2x2=t, or, -3-4xdx=dtI1=tdt=23t32+c1I1=234-3x-2x232+c1

I2=4-3x-2x2dx=22-32x-x2dx=2174-94-32x-x2dx=21722-94+32x+x2dx=21722-x+322dx=2sinx+32172+c2=2sin2x+317+c2

Using (i), we get

I=-34×234-3x-2x232-54×2sin2x+317+C I=-124-3x-2x232-524sin2x+317+C

Page No 19.159:

Question 14:

(2x+5)10-4x-3x2dx

Answer:

I=2x+510-4x-3x2dxLet 2x+5=Addx10-4x-3x2+B2x+5=A-4-6x+B2x+5=-6Ax+B-4A2=-6A and B-4A=5A=-13 and B=113

2x+5=-13-4-6x+113I=-13-4-6x10-4x-3x2dx+11310-4x-3x2dxLet I=-13I1+113I2               ...iNow,I1=-4-6x10-4x-3x2dxLet 10-4x-3x2=t, or, -4-6xdx=dtI1=tdt=23t32+c1I1=2310-4x-3x232+c1

I2=10-4x-3x2dx=3103-43x-x2dx=3269-49-43x-x2dx=32632-49+43x+x2dx=32632-x+232dx=3sinx+23263+c2=3sin3x+226+c2

Using (i), we get

I=-13×2310-4x-3x232+113×3sin3x+226+C I=-2910-4x-3x232+1133sin3x+226+C



Page No 19.176:

Question 1:

2x+1x+1 x-2 dx

Answer:

2x+1x+1x-2 dx       Let 2x+1x+1x-2=Ax+1+Bx-2      ....(1)2x+1x+1x-2=Ax-2+Bx+1x+1x-2Then ,2x+1=Ax-2+Bx+1             ....(2)Putting x-2=0 or, x=2 in eq (2) 2×2+1=A×0+B2+1B=53Putting x+1=0 or, x=-1 in eq (2) 2×-1+1+A-1-2+B×0-1=A-3A=13Substituting the values of A and B in eq  (1) , we get  2x+1x+1x-2=13x+1+53x-22x+1dxx+1x-2=131x+1dx+531x-2dx                            =13 ln x+1+53 ln x-2+C

Page No 19.176:

Question 2:

1xx-2 x-4 dx

Answer:

1xx-2x-4dxLet 1xx-2x-4=Ax+Bx-2+Cx-41xx-2x-4=Ax-2x-4+Bxx-4+Cx·x-2xx-2x-41=Ax-2x-4+Bx·x-4+Cx.x-2       ...(1)Putting  x=0 in eq (1) 1=A0-20-4+B×0+C×018=APutting x-2=0 or  x=2 in eq (1)1=A×0+B22-4+C×2×0B=-14Putting x-4=0 or  x=4 in eq (1)1=A×0+B×0+C·44-2C=181xx-2x-4=18x-14x-2+18x-4dxxx-2x-4=181xdx-141x-2dx+181x-4dx                                    =18 ln x-14 ln x-2+18 ln x-4+C                                    =18ln x+ln x-4-2 ln x-2+C                                    =18ln xx-4x-22+C

Page No 19.176:

Question 3:

x2+x-1x2+x-6 dx

Answer:

x2+x-1x2+x-6dx=x2+x-6+6-1x2+x-6dx=x2+x-6x2+x-6dx+51x2+x-6dx=dx+51x2+3x-2x-6dx=dx+51xx+3-2x+3dx=dx+51x-2x+3dx                            . . . (1)Let1x-2x+3=Ax-2+Bx+31x-2x+3=Ax+3+Bx-2x-2x+31=Ax+3+Bx-2               ....(2) Putting x+3=0 or x=-3 in eq (2)1=A×0+B-3-2B=-15Putting x-2=0 or x=2 in eq (2)1=A2+3+B×0A=15

1x-2x+3=15x-2-15x+31x-2x+3dx=15dxx-2-15dxx+3                                      =15 ln x-2-15 ln x+3+C                                      =15 ln x-2x+3+C                                 . . . (3)From eq (1) & eq (3)x2+x-1x2+x-6dx=x+55 ln x-2x+3+C                                   =x+ln x-2-ln x+3+C

Page No 19.176:

Question 4:

3+4x-x2x+2 x-1 dx

Answer:

3+4x-x2x2-x+2x-2dx=-x2+4x+3x2+x-2dx-x2+4x+3x2+x-2=-1+5x+1x2+x-2                      . . . (1)5x+1x2+x-2=5x+1x2+2x-x-2                        =5x+1xx+2-1x+2Let5x+1x-1x+2=Ax-1+Bx+25x+1x-1x+2=Ax+2+Bx-1x-1x+25x+1=Ax+2+Bx-1                ...(2)Putting x+2=0 or x=-2 in eq (2)5x-2+1=A×0+B-2-1B=3Putting x-1=0  or x=1 in eq (2) 5×1+1=A3+B×0A=25x+1x-1x+2=2x-1+3x+2                            . . . (3)From (1) & (3)-x2+4x+3x2+x-2=-1+2x-1+3x+2-x2+4x+3x2+x-2dx=-1 dx+2x-1dx+3x+2 dx                                      =-x+2 lnx-1+3 lnx+2+C  

Page No 19.176:

Question 5:

x2+1x2-1 dx

Answer:

x2+1x2-1dx=x2-1+2x2-1dx=dx+21x2-12dx=dx+21x-1 x+1dx               ... 1 1x-1x+1=Ax-1+Bx+11x-1 x+1=A x+1+Bx-1x-1 x+11=A x+1 B x-1       ....(2)Putting x+1=0 or x=-1 in eq (2)1=A×0+B -1 -1B=-12Putting x-1=0 or x=1 in eq (2)1=A 1+1+B×0A=12 1x-1x+1=12x-1-12x+1        ...(3)From eq (1) & eq (3) x2+1x2-1dx=dx+212 x-1-12 x+1dx                        =dx+dxx-1-dxx+1                        =x+ln x-1=-ln x+1+C                        =x+ln x-1x+1+C

Page No 19.176:

Question 6:

x2x-1 x-2 x-3 dx

Answer:

 x2 x-1 x-2 x-3dxLetx2x-1 x-2 x-3=Ax-1+Bx-2+Cx-3x2x-1 x-2 x-3=A x-2 x-3+B x-1 x-3+C x-1 x-2x-1 x-2 x-3x2=A x-2 x-3+B x-1 x-3+C x-1 x-2         ....(1)Putting x-1=0 or x=1 in eq (1)1=A 1-2 1-31=A -1 -2A=12Putting x-2=0 or x=2 in eq (1) 4=B 2-1 2-3B=-4Putting x-3=0 or x=3 in eq (1) 9=C 3-1 3-2C=92x2x-1 x-2 x-3=12 x-1-4x-2+92 x-3x2 x-1 x-2 x-3dx=121x-1dx-41x-2dx+921x-3dx                                             =12ln x-1-4 ln x-2+92 lnx-3+C

Page No 19.176:

Question 7:

5xx+1 x2-4 dx

Answer:

 5xx+1 x2-4dxLet5xx+1 x-2 x+2=Ax+1+Bx-2+Cx+25xx+1 x-2 x+2=A x-2 x+2+B x+1 x+2+C x+1 x-2x+1 x-2 x+25x=A x-2 x+2+B x+1 x+2+C x+1 x-2       ....(1)Putting x-2=0 or x=2 in eq (1)5×2=B 2+1 2+2B=103×4        =56Putting x+2=0 or x=-2 in eq (1)5×-2=C -2+1 -2-2-10-1×-4 =CC=-52Putting x+1=0 or x=-1 in eq (1) -5=A -1-2 -1+2-5-3=AA=53 5xx+1 x-2 x+2=53×1x+1+56 x-2- 52 x+25xx+1 x-2 x+2=56×2x+1+56 x-2-56 3x+2 5x x+1 x-2 x+2dx=562x+1 dx+561x-2dx-563x+2 dx                                                   =562 ln x+1+ln x-2-3 ln x+2+C                                                   =56 ln x+12+ln x-2-ln x+23+C                                                   =56 ln x+12 x-2x+23+C

Page No 19.176:

Question 8:

x2+1xx2-1 dx

Answer:

x2+1x x2-1dx=x2+1x x-1 x+1dxLetx2+1x x-1 x+1=Ax+Bx-1+Cx+1x2+1x x-1 x+1=A x-1 x+1+B x x+1+C x x-1x x-1 x+1x2+1=A x-1 x+1+B x x+1+C x x-1        ....(1)Putting x-1=0 or x=1 in eq (1) 1+1=A×0+B 1 1+1+C×0B=1Putting x=0 in eq (1) 0+1=A 0-1 0+1A=-1Putting x+1=0 or x=-1 in eq (1) -12+1=A×0+B×0+C-1 -1-12=C×2C=1x2+1x x2-1=-1x+1x-1+1x+1x2+1x x2-1dx=-1xdx+1x-1dx+1x+1dx                                =-ln x+ln x-1+ln x+1+C                                =-ln x+ln x2-1+C                                =ln x2-1x+C

Page No 19.176:

Question 9:

2x-3x2-1 2x+3 dx

Answer:

 2x-3 x2-1 2x+3dx=2x-3 x-1 x+1 2x+3dxLet 2x-3x-1 x+1 2x+3=Ax-1+Bx+1+C2x+32x-3x-1 x+1 2x+3=A x+1 2x+3+B x+1 2x+3+C x2-1x-1 x+1 2x+32x-3=A x+1 2x+3+B x-1 2x+3+C x+1 x-1        ...(1)Putting x+1=0 or x=-1 in eq (1)-2-3=B -1-1 -2+3-5=B -2 1B=52Putting x-1=0 or x=1 in eq (1)2-3=A 1+1 2+3-1=A 2 5A=-110Putting 2x+3=0 or x=-32in eq (1) 2×-32-3=A×0+B×0+C-32+1 -32-1-6=C -12 -52C=-2452x-3x-1 x+1 2x+3=-110 x-1+52 x+1-245 2x+32x-3x-1 x+1 2x+3 dx=-1101x-1dx+521x+1dx-24512x+3dx                                                      =-110 ln x-1+52 ln x+1-245 ln 2x+33+C                                                      =-110 ln x-1+52 ln x+1-125 ln 2x+3+C

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Question 10:

x3x-1 x-2 x-3 dx

Answer:

x3 x-1 x-2x-3dx=x3 x-1 x2-5x+6dx=x3 x3-5x2+6x-x2+5x-6dx=x3 x3-6x2+11x-6dx                       x3x3-6x2+11x-6=1+6x2+11x+6x2-6x2+11x-6x3x3-6x2+11x-6=1+6x2-11x+6x-1 x-2 x-3x3x-1 x-2 x-3 dx=dx+6x2-11x+6x-1 x-2 x-3dx          ...1

Let6x2-11x+6x-1 x-2 x-3=Ax-1+Bx-2+Cx-36x2-11x+6x-1 x-2 x-3=A x-2 x-3+B x-1 x-3+C x-1 x-2x-1 x-2 x-36x2-11x+6=A x-2 x-3+B x-1 x-3+C x-1 x-2          ...(2)Putting x-2=0 or x=2 in eq (2) 6×4-22+6=B 2-1 2-38=B -1B=-8Putting x-3=0 or x=3 in eq (2)6×32-11×3+6=C 3-1 3-227=C 2 1C=272Putting x-1=0 or x=1 in eq (2) 6×1-11+6=A 1-2 1-31=A -1 -2A=126x2-11x+6x-1 x-2 x-3=12x-1-8x-2+272x-3           ...(3)From eq (2) and eq (3)x3 dxx-1 x-2 x-3=dx+121x-1dx-81x-2dx+2721x-3dx                                             =x+12 ln x-1-8 ln x-2+272 ln x-3+C

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Question 11:

sin 2x1+sin x 2+sin x dx

Answer:

We have, I=sin 2x dx1+sin x 2+sin x=2 sin x cos x dx1+sin x 2+sin xPutting sin x=tcos x dx=dtI=2t dt1+t 2+t=2t dt1+t 2+tLet t1+t 2+t=A1+t+B2+tt1+t 2+t=A 2+t+B 1+t1+t 2+tt=A 2+t+B 1+tPutting 2+t=0t=-2-2=A×0+B -2+1-2=B -1B=2let t+1=0t=-1-1=A -1+2+B×0A=-1 I=2-1t+1+2t+2dt=2 -log t+1+2 log t+2+C=4 log t+2-2 log t+1+C=log t+24t+12+C=log sin x+24sin x+12+C

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Question 12:

2xx2+1 x2+3 dx

Answer:

We have,I=2x dxx2+1 x2+3Putting x2=t2x dx=dtI=dtt+1 t+3Let 1t+1 t+3=At+1+Bt+31t+1 t+3=A t+3+B t+1t+1 t+31=A t+3+B t+1Putting t+3=0t=-31=A×0+B -3+1B=-12Putting t+1=0t=-11=A -1+3+B -1+11=A×2+B×0A=12Then,I=12dtt+1-12dtt+3=12 log t+1-12 log t+3+C=12 log t+1t+3+C=12 log x2+1x2+3+C

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Question 13:

1x log x 2+log x dx

Answer:

We have,I=dxx log x2+log xPutting log x=t1x dx=dtI=dtt t+2Let 1t t+2=At+Bt+21t t+2=At+2+Btt t+21=A t+2+BtPutting t+2=0t=-21=A×0+B -2B=-12Putting t=01=A 0+2+B ×0A=12Then,I=12dtt-12dtt+2=12 log t-log t+2+C=12 log tt+2+C=12 log log xlogx+2+C

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Question 14:

Evaluate the following integrals:

x2+x+1x2+1x+2dx

Answer:

Let I=x2+x+1x2+1x+2dxWe expressx2+x+1x2+1x+2=Ax+2+Bx+Cx2+1x2+x+1=Ax2+1+Bx+Cx+2Equating the coefficients of x2, x and constants, we get1=A+B      and      1=2B+C      and     1=A+2Cor A=35      and      B=25      and      C=15 I=35x+2+25x+15x2+1dx     =351x+2dx+25xx2+1 dx+151x2+1 dx     =35I1+25I2+15I3       ...(1)Now, I1=1x+2dx      Let x+2=u      On differentiating both sides, we get      dx=du I1=1udu       =logu+c1       =logx+2+c1      ...(2)And, I2=xx2+1 dx      Let x2+1=u      On differentiating both sides, we get      2x dx=du I2=121udu       =12logu+c2       =12logx2+1+c2      ...(3)And, I3=1x2+1 dx          =tan-1x+c3       ...(4)From (1), (2), (3) and (4), we get I=35logx+2+c1+2512logx2+1+c2+15tan-1x+c3      =35logx+2+15logx2+1+15tan-1x+cHence, x2+x+1x2+1x+2dx=35logx+2+15logx2+1+15tan-1x+c

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Question 15:

ax2+bx+cx-a x-b x-c dx, where a, b, c are distinct

Answer:

We have,I=ax2+bx+cx-a x-b x-c dxLet ax2+bx+cx-a x-b x-c=Ax-a+Bx-b+Cx-cax2+bx+c=Ax-b x-c+B x-cx-a+Cx-a x-bax2+bx+c=Ax2-b+cx+bc+Bx2-c+ax+ca+Cx2-a+bx+abax2+bx+c=A+B+Cx2-Ab+c+Bc+a+Ca+bx+Abc+Bca+CabEquating the coefficients on both sides, we geta=A+B+C                                                  .....1b=-Ab+c+Bc+a+Ca+b              .....2c=Abc+Bca+Cab                                        .....3Solving 1, 2 and 3, we getA=a3+ab+ca-ba-cB=ab2+b2+cb-ab-cC=ac2+bc+cc-ac-bI=a3+ab+ca-ba-c×1x-a+ab2+b2+cb-ab-c×1x-b+ac2+bc+cc-ac-b×1x-c dx=a3+ab+ca-ba-clog x-a+ab2+b2+cb-ab-clog x-b+ac2+bc+cc-ac-blog x-c+K

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Question 16:

Evaluate the following integrals:

xx2+1x-1dx

Answer:

Let I=xx2+1x-1dxWe expressxx2+1x-1=Ax-1+Bx+Cx2+1x=Ax2+1+Bx+Cx-1Equating the coefficients of x2, x and constants, we get0=A+B      and      1=-B+C      and     0=A-Cor A=12      and      B=-12      and      C=12 I=12x-1+-12x+12x2+1dx     =121x-1dx-12xx2+1 dx+121x2+1 dx     =12I1-12I2+12I3       ...(1)Now, I1=1x-1dx      Let x-1=u      On differentiating both sides, we get      dx=du I1=1udu       =logu+c1       =logx-1+c1      ...(2)And, I2=xx2+1 dx      Let x2+1=u      On differentiating both sides, we get      2x dx=du I2=121udu       =12logu+c2       =12logx2+1+c2      ...(3)And, I3=1x2+1 dx          =tan-1x+c3       ...(4)From (1), (2), (3) and (4), we get I=12logx-1+c1-1212logx2+1+c2+12tan-1x+c3      =12logx-1-14logx2+1+12tan-1x+cHence, xx2+1x-1dx=12logx-1-14logx2+1+12tan-1x+c

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Question 17:

1x-1 x+1 x+2 dx

Answer:

We have,I=dxx-1 x+1 x+2Let 1x-1 x+1 x+2=Ax-1+Bx+1+Cx+21x-1 x+1 x+2=A x+1 x+2+B x-1 x+2+C x-1 x+1x-1 x+1 x+21=A x+1 x+2+B x-1 x+2+C x-1 x+1Putting x-1=0x=11=A 1+1 1+2+B×0+C×01=A×6A=16Putting x+1=0x=-11=A×0+B -2 1+C×0B=-12Putting x+2=0x=-21=A×0+B×0+C -2-1 -2+11=C×3C=13I=16dxx-1-12dxx+1+13dxx+2=16 log x-1-12 log x+1+13 log x+2+C=16 log x-1-36 log x+1+26log x+2+C=16 log x-1-3 log x+1+2 log x+2+C=16log x-1 x+22x+13 +C

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Question 18:

Evaluate the following integrals:

x2x2+4x2+9dx

Answer:

Let I=x2x2+4x2+9dxWe expressx2x2+4x2+9=Ax+Bx2+4+Cx+Dx2+9x2=Ax+Bx2+9+Cx+Dx2+4Equating the coefficients of x3, x2, x and constants, we get0=A+C      and      1=B+D      and      0=9A+4C      and     0=9B+4Dor A=0      and      B=-45      and      C=0      and      D=95I=-45x2+4+95x2+9dx     =-451x2+4dx+951x2+9 dx     =-45×12tan-1x2+95×13tan-1x3+c     =-25tan-1x2+35tan-1x3+cHence, x2x2+4x2+9dx=-25tan-1x2+35tan-1x3+c

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Question 19:

5x2-1x x-1 x+1 dx

Answer:

We have,I=5x2-1 dxx x-1 x+1Let 5x2-1x x-1 x+1=Ax+Bx-1+Cx+15x2-1x x-1 x+1=A x2-1 +Bx·x+1+C·x· x-1x x-1 x+15x2-1=A x2-1+B·x x+1+C·x· x-1Putting  x=15-1=A×0+B 1 1+1+C×04=B 2B=2Putting  x=05×0-1=A 0-1+B×0+C×0-1=A -1A=1Putting  x+1=0x=-15-1=A×0+B×0+C -1 -2C=2I=dxx+2dxx-1+2dxx+1=log x+2 log x-1+2 log x+1+C=log x+2 log x2-1+C=log xx2-12+C

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Question 20:

x2+6x-8x3-4x dx

Answer:

We have,I=x2+6x-8x3-4xdx=x2+6x-8x x2-4dx=x2+6x-8x x-2 x+2dxLet x2+6x-8x x-2 x+2=Ax+Bx-2+Cx+2x2+6x-8x x-2 x+2=A x-2 x+2+B x x+2+C x x-2x x-2 x+2x2+6x-8=A x2-4+B x2+2x+C x2-2xPutting  x-2=0x=24+6×2-8=A×0+B 4+48=B×8B=1Putting  x=-24-12-8=A×0+B×0+C×8C=-2Putting  x=0-8=A -4+B×0+C×0A=2I=2x+dxx-2-2dxx+2=2 log x+log x-2-2 log x+2+C=log x2+log x-2-log x+22+C=log x2x-2x+22+C

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Question 21:

x2+12x+1 x2-1 dx

Answer:

We have,I=x2+1 dx2x+1 x2-1=x2+1 dx2x+1 x-1 x+1Let x2+12x+1 x-1 x+1=A2x+1+Bx-1+Cx+1x2+12x+1 x-1 x+1=A x2-1+B 2x+1 x+1+C 2x+1 x-12x+1 x-1 x+1x2+1=A x2-1+B 2x+1 x+1+C 2x+1 x-1Putting x-1=0x=11+1=A×0+B 2+1 1+1+C×02=B32B=13Putting x+1=0x=-11+1=A×0+B×0+C -2+1 -1-12=C -1 -2C=1Putting 2x+1=0x=-12-122+1=A 14-114+1=A -3454=A -34A=-53I=-53dx2x+1+13dxx-1+dxx+1=-53× log 2x+12+13 log x-1+log x+1+C=-56 log 2x+1+13 log x-1+log x+1+C



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Question 22:

1x6 log x2+7 log x+2 dx

Answer:

We have,I=dxx 6 log x2+7 log x+2Putting log x=t1x dx=dtI=dt6t2+7t+2=dt3t+2 2t+1Let 13t+2 2t+1=A3t+2+B2t+113t+2 2t+1=A 2t+1+B 3t+23t+2 2t+11=A 2t+1+B 3t+2Putting 2t+1=0t=-121=0+B 3×-12+21=B 12B=2Putting 3t+2=0t=-231=A 2×-23+1+01=A -43+11=A -13A=-3I=-33t+2+22t+1dt=-3 log 3t+23+2 log 2t+12+C=-log 3t+2+log 2t+1+C=log 2t+13t+2+C=log 2 log x+13 log x+2+C

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Question 23:

1xxn+1 dx

Answer:

We have,I=dxx xn+1=xn-1dxxn-1x xn+1=xn-1 dxxn xn+1Putting xn=tn xn-1 dx=dtxn-1dx=dtnI=1ndtt t+1Let 1t t+1=At+Bt+11t t+1=A t+1+Btt t+11=A t+1+BtPutting t+1=0t=-11=A×0+B -1B=-1Putting t=01=A 0+1+B×0A=1Then,I=1ndtt-1ndtt+1=1n log t-1nlog t+1+C=1n log tt+1+C=1n log xnxn+1+C

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Question 24:

xx2-a2 x2-b2 dx

Answer:

We have,I=x dxx2-a2 x2-b2Putting x2=t2x dx=dtx dx=dt2I=12dtt-a2 t-b2Let 1t-a2 t-b2=At-a2+Bt-b21t-a2 t-b2=A t-b2+B t-a2t-a2 t-b21=A t-b2+B t-a2Putting t=b21=A×0+B b2-a2B=1b2-a2Putting t=a21=A a2-b2+B ×0A=1a2-b2I=12dtt-a2 t-b2=12 a2-b2dtt-a2+12 b2-a2dtt-b2=12 a2-b2 log t-a2+12 b2-a2 log t-b2+C=12 a2-b2 log t-a2-log t-b2+C=12 a2-b2 log t-a2t-b2 +C=12 a2-b2 log x2-a2x2-b2+C

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Question 25:

Evaluate the following integrals:

x2+1x2+4x2+25dx

Answer:

Let I=x2+1x2+4x2+25dxWe expressx2+1x2+4x2+25=Ax+Bx2+4+Cx+Dx2+25x2+1=Ax+Bx2+25+Cx+Dx2+4Equating the coefficients of x3, x2, x and constants, we get0=A+C      and      1=B+D      and      0=25A+4C      and     1=25B+4Dor A=0      and      B=-17      and      C=0      and      D=87I=-17x2+4+87x2+25dx     =-171x2+4dx+871x2+25 dx     =-17×12tan-1x2+87×15tan-1x5+c     =-114tan-1x2+835tan-1x5+cHence, x2+1x2+4x2+25dx=-114tan-1x2+835tan-1x5+cA

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Question 26:

Evaluate the following integrals:

x3+x+1x2-1dxx3+x+1x2-1dx

Answer:

Let I=x3+x+1x2-1dxHere the integrand x3+x+1x2-1 is not a proper rational function, so we divide x3+x+1 by x2-1 and find thatx3+x+1x2-1=x+2x+1x2-1=x+2x+1x+1x-1Let 2x+1x+1x-1=Ax+1+Bx-12x+1=Ax-1+Bx+1Equating the coefficients of x and constants, we get2=A+B      and     1=-A+Bor A=12      and      B=32I=x+12x+1+32x-1dx     =x dx+121x+1dx+321x-1 dx     =x22+12logx+1+32logx-1+c     =x22+12logx+1+32logx-1+cHence, x3+x+1x2-1dx=x22+12logx+1+32logx-1+c

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Question 27:

Evaluate the following integrals:

3x-2x+12x+3dx
 

Answer:

Let I=3x-2x+12x+3dxWe express3x-2x+12x+3=Ax+1+Bx+12+Cx+33x-2=Ax+1x+3+Bx+3+Cx+12Equating the coefficients of x2, x and constants, we get0=A+C      and      3=4A+B+2C      and      -2=3A+3B+Cor A=114      and      B=-52      and      C=-114I=114x+1+-52x+12+-114x+3dx     =1141x+1dx-521x+12 dx-1141x+3 dx     =114logx+1+52x+1-114logx+3+cHence, 3x-2x+12x+3dx=114logx+1+52x+1-114logx+3+c

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Question 28:

2x+1x+2 x-32 dx

Answer:

We have,I=2x+1 dxx+2 x-32Let 2x+1x+2 x-32=Ax+2+Bx-3+Cx-322x+1x+2 x-32=A x-32+B x+2 x-3+C x+2x+2 x-322x+1=A x2-6x+9+B x2-x-6+C x+22x+1=A+B x2+-6A-B+C x+9A-6B+2CEquating the coefficients of like termsA+B=0                           .....  1-6A-B+C=2              .....  29A-6B+2C=1              .....3Solving 1, 2 and 3, we getA=-325, B=325 and C=752x+1 dxx+2 x-32=-325 x+2+325 x-3+75 x-32I=-325dxx+2+325dxx-3+75x-3-2 dx=-325 log x+2+325 log x-3+75x-3-1-1+C=-325log x+2+325 log x-3-75 x-3+C

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Question 29:

x2+1x-22 x+3 dx

Answer:

We have,I=x2+1 dxx-22 x+3Let x2+1x-22 x+3=Ax-2+Bx-22+Cx+3x2+1x-22 x+3=A x-2 x+3+B x+3+C x-22x-22 x+3x2+1=A x2-2x+3x-6+B x+3+C x2-4x+4x2+1=A x2+x-6 +B x+3+C x2-4x+4Equating coefficients of like termsA+C=1                        .....   1A+B-4C=0                    .....   2-6A+3B+4C=1              .....   3Solving 1, 2 and 3, we getA=35, B=1 and C=25I=35dxx-2+dxx-22+25dxx+3=35 log x-2+x-2-2+1-2+1+25 log x+3+C=35log x-2-1x-2+25 log x+3+C

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Question 30:

xx-12 x+2 dx

Answer:

We have,I=x dxx-12 x+2Let xx-12 x+2=Ax-1+Bx-12+Cx+2xx-12 x+2=A x-1 x+2+B x+2+C x-12x-12 x+2x=A x2+2x-x-2+B x+2+C x2-2x+1x=A x2+x-2 +B x+2+C x2-2x+1x=A+C x2+A+B-2C x+-2A+2B+CEquating coefficients of like termsA+C=0          .....1A+B-2C=1          .....2-2A+2B+C=0          .....3Solving 1, 2 and 3, we getA=29, B=13 and C=-29xx-12 x+2=29 x-1+13 x-12-29 x+2I=29dxx-1+13dxx-12-29dxx+2=29 log x-1+13×-1x-1-29 log x+2+C=29log x-1x+2-13 x-1+C

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Question 31:

x2x-1 x+12 dx

Answer:

We have,I=x2 dxx-1 x+12Let x2x-1 x+12=Ax-1+Bx+1+Cx+12x2x-1 x+12=A x+12+B x+1 x-1+C x-1x+12 x-1x2=A x2+2x+1+B x2-1+C x-1x2=A+B x2+x 2A+C +A-B-CEquating coefficients of like termsA+B=1  .....12A+C=0  .....2A-B-C=0  .....3Solving 1, 2  and 3, we getA=14, B=34 and C=-12x2x-1 x+12=14 x-1+34 x+1-12 x+12I=14dxx-1+34dxx+1-12dxx+12=14 log x-1+34 log x+1-12×-1x+1+C=14log x-1+34 log x+1+12 x+1+C

Page No 19.177:

Question 32:

x2+x-1x+12 x+2 dx

Answer:

We have,I=x2+x-1 dxx+12 x+2Let x2+x-1x+12 x+2=Ax+1+Bx+12+Cx+2x2+x-1x+12 x+1=A x+1 x+2+B x+2+C x+12x+12 x+2x2+x-1=A x2+3x+2+B x+2+C x2+2x+1Equating coefficients of like termsA+C=1             .....13A+B+2C=1     .....22A+2B+C=-1  .....3Solving 1, 2 and 3, we getA=0 B=-1C=1x2+x-1x+12 x+2=-1x+12+1x+2I=-dxx+12+dxx+2=-x+1-2dx+dxx+2=-x+1-2+1-2+1 +log x+2+C=1x+1+log x+2+C

Page No 19.177:

Question 33:

2x2+7x-3x2 2x+1 dx

Answer:

We have,I=2x2+7x-3 dxx2 2x+1Let 2x2+7x-3x2 2x+1=Ax+Bx2+C2x+12x2+7x-3x2 2x+1=A x 2x+1+B 2x+1+Cx2x2 2x+12x2+7x-3=A 2x2+x+B 2x+1+Cx22x2+7x-3=2A+C x2+A+2Bx+BEquating coefficients of like terms2A+C=2     .....1A+2B=7     .....2B=-3     .....3Solving 1,  2 and 3, we getA=13B=-3C=-242x2+7x-3x2 2x+1=13x-3x2-242x+1I=13dxx-3x-2 dx-24dx2x+1=13 log x+3x-24 log 2x+12+C=13 log x+3x-12 log 2x+1+C

Page No 19.177:

Question 34:

5x2+20x+6x3+2x2+x dx

Answer:

We have,I=5x2+20x+6x3+2x2+x=5x2+20x+6 dxx x2+2x+1=5x2+20x+6 dxx x+12Let 5x2+20x+6x x+12=Ax+Bx+1+Cx+125x2+20x+6x x+12=A x+12 +B x x+1+C xx x+125x2+20x+6=A x2+2x+1+B x2+x+Cx5x2+20x+6=A+B x2+2A+B+C x+AEquating coefficients of like termsA+B=5                 .....12A+B+C=20     .....2 A=6                     .....3Solving 1, 2 and 3, we getA=6  B=-1C=95x2+20x+6x x+12=6x-1x+1+9x+12I=6dxx-dxx+1+9dxx+12=6 log x-log x+1-9x+1+C

Page No 19.177:

Question 35:

18x+2 x2+4 dx

Answer:

We have,I=18x+2 x2+4 dxLet 18x+2 x2+4=Ax+2+Bx+Cx2+418x+2 x2+4=A x2+4+Bx+C x+2x+2 x2+418=Ax2+4A+Bx2+2Bx+Cx +2C18=A+B x2+x 2B+C+4A+2CEquating coefficients of like termsA+B=0           .....12B+C=0        .....24A+2C=18   .....3Solving 1, 2 and 3, we getA=94B=-94C=9218x+2 x2+4=94 x+2+-94x+92x2+418x+2 x2+4=94 x+2-94 xx2+4+92 x2+418 dxx+2 x2+4=94dxx+2-94x dxx2+4+92dxx2+22let x2+4=t2xdx=dtx dx=dt2I=94dxx+2-98dtt+92dxx2+22=94 log x+2-98 log t+92×12 tan-1 x2+C'=94 log x+2-98 log x2+4+94 tan-1 x2+C'

Page No 19.177:

Question 36:

5x2+1 x+2 dx

Answer:

We have,I=5 dxx2+1 x+2 Let 5x+2 x2+1=Ax+2+Bx+Cx2+15x+2 x2+1=A x2+1+Bx+C x+2x+2 x2+15=A x2+1+Bx2+2Bx+Cx +2C5=A+B x2+2B+C x+A+2CEquating coefficients of like termsA+B=0     .....12B+C=0     .....2A+2C=5     .....3Solving 1, 2 and 3, we getA=1B=-1C=25x+2 x2+1=1x+2+-x+2x2+15 dxx+2 x2+1=dxx+2-x dxx2+1+2dxx2+1let x2+1=t2xdx=dtx dx=dt2I=dxx+2-12dtt+2dxx2+12=log x+2-12 log t+2 tan-1x+C'=log x+2-12 log x2+1+2 tan-1x+C'

Page No 19.177:

Question 37:

xx+1 x2+1 dx

Answer:

We have,I=x dxx+1 x2+1 Let xx+1 x2+1=Ax+1+Bx+Cx2+1xx+1 x2+1=A x2+1+Bx+C x+1x+1 x2+1x=A x2+1+Bx2+Bx+Cx+Cx=A+B x2+B+C x+A+CEquating coefficients of like termsA+B=0     .....1B+C=1     .....2A+C=0     .....3Solving 1, 2 and 3, we getA=-12B=12C=12xx+1 x2+1=-12 x+1+x2+12x2+1x dxx+1 x2+1=-12dxx+1+12x dxx2+1+12dxx2+1let x2+1=t2x dx=dtx dx=dt2I=-12dxx+1+14dtt+12dxx2+12=-12 log x+1+14 log t+12 tan-1x+C'=-12 log x+1+14 log x2+1+12 tan-1x+C'

Page No 19.177:

Question 38:

11+x+x2+x3 dx

Answer:

We have,I=dx1+x+x2+x3 =dx1+x+x2 1+x=dxx+1 x2+1Let 1x+1 x2+1=Ax+1+Bx+Cx2+11x+1 x2+1=A x2+1+Bx+C x+1x+1 x2+11=A x2+1+Bx2+Bx+Cx+C1=A+B x2+B+C x+A+CEquating coefficients of like termsA+B=0     .....1B+C=0     .....2A+C=1     .....3Solving 1, 2 and 3, we getA=12B=-12C=12I=12dxx+1+12-x+1x2+1 dx=12dxx+1-12x dxx2+1+12dxx2+12let x2+1=t2x dx=dtx dx=dt2I=12dxx+1-14dtt+12dxx2+12=12 log x+1-14 log t+12 tan-1 x+C'=12 log x+1-14 log x2+1+12 tan-1 x+C'

Page No 19.177:

Question 39:

1x+12 x2+1 dx

Answer:

We have,I=dxx+12 x2+1 Let 1x+12 x2+1=Ax+1+Bx+12+Cx+Dx2+11x+12 x2+1=A x+1 x2+1+B x2+1+Cx+D x+12x+12 x2+11=A x3+x+x2+1+B x2+1+Cx+Dx2+2x+11=A x3+x2+x+1+B x2+1+Cx3+2Cx2+Cx+Dx2+2Dx+D1=A+C x3+A+B+2C+Dx2+A+C+2D x+A+B+DEquating coefficients of like termsA+C=0                     .....1A+B+2C+D=0     .....2A+C+2D=0            .....3A+B+D=1             .....4A=12, B=12, C=-12 and D=01x+12 x2+1=12 x+1+12 x+12-12 ×xx2+1dxx+12 x2+1=12dxx+1+12dxx+12-12x dxx2+1Putting x2+1=t2x dx=dtx dx=dt2I=12dxx+1+12dxx+12-14dtt=12 log x+1-12 x+1-14 log t+C'=12 log x+1-12 x+1 -14 log x2+1+C'

Page No 19.177:

Question 40:

2xx3-1 dx

Answer:

We have,I=2x dxx3-1 =2x dxx-1 x2+x+1Let 2xx-1 x2+x+1=Ax-1+Bx+Cx2+x+12xx-1 x2+x+1=A x2+x+1+Bx+C x-1x-1 x2+x+12x=A x2+x+1+Bx2-Bx+Cx-C2x=A+B x2+A-B+C x+A-CEquating coefficients of like termsA+B=0           .....1A-B+C=2      .....2A-C=0           .....3Solving 1, 2 and 3, we getA=23B=-23C=23I=23dxx-1+23-x+1x2+x+1dxLet -x+1=addx x2+x+1+b-x+1=a 2x+1+b-x+1=2a x+a+bEquating coefficients of like terms 2a=-1a=-12Anda+b=1-12+b=1b=32I=23dxx-1+23-12 2x+1+32x2+x+1dx=23dxx-1-132x+1x2+x+1 dx+dxx2+x+1=23dxx-1-132x+1x2+x+1dx+dxx2+x+14-14+1=23dxx-1-132x+1 dxx2+x+1+dxx+122+322let x2+x+1=t2x+1 dx=dtThen,I=23dxx-1-13dtt+dxx+122+322=23 log x-1-13 log t+23 tan-1 x+1232+C'=23 log x-1-13 log x2+x+1+23 tan-1 2x+13+C'

Page No 19.177:

Question 41:

dxx2+1 x2+4

Answer:

We have,I=dxx2+1 x2+4 Putting x2=tThen, 1x2+1 x2+4 =1t+1 t+4Let 1t+1 t+4=At+1+Bt+41=A t+4+B t+1Putting t+4=0t=-41=A×0+B -3B=-13Putting t+1=0t=-11=A -1+4+B×0A=131t+1 t+4=13 t+1-13 t+41x2+1 x2+4=13 x2+1-13 x2+22dxx2+1 x2+4=13dxx2+12-13dxx2+22=13 tan-1x-13×12 tan-1x2+C=13 tan-1x-16 tan-1 x2+C

Page No 19.177:

Question 42:

x2x2+1 3x2+4 dx

Answer:

We have,I=x2 dxx2+1 3x2+4 Putting x2=tThen, x2x2+1 3x2+4 =tt+1 3t+4Let tt+1 3t+4=At+1+B3t+4tt+1 3t+4=A 3t+4+B t+1 t+1 3t+4t=A 3t+4+B t+1Putting t+1=0t=-1-1=A -3+4+0A=-1Putting 3t+4=0t=-43-43=0+B -43+1-43=B×-13B=4tt+1 3t+4=-1t+1+43t+4x2x2+1 3x2+4=-1x2+1+43x2+4x2x2+1 3x2+4=-1x2+1+43 x2+43x2 dxx2+1 3x2+4=-dxx2+1+43dxx2+232=-tan-1 x+43×32 tan-1 3x2+C=-tan-1 x+23 tan-1 3x2+C

Page No 19.177:

Question 43:

3x+5x3-x2-x+1 dx

Answer:

We have,I=3x+5dxx3-x2-x+1=3x+5dxx2x-1-1x-1=3x+5dxx2-1 x-1=3x+5dxx-1 x+1 x-1=3x+5dxx-12 x+1Let 3x+5x-12 x+1=Ax+1+Bx-1+Cx-123x+5x-12 x+1=Ax-12+Bx+1 x-1+Cx+1x+1 x-123x+5=Ax2-2x+1+Bx2-1+Cx+C3x+5=A+Bx2+-2A+Cx+A-B+CEquating coefficient of like termsA+B=0            .....1-2A+C=3      .....2A-B+C=5     .....3Solving these three equations we getA=12B=-12C=43x+5x-12 x+1=12x+1-12x-1+4x-12I=12dxx+1-12dxx-1+4x-1-2 dx=12log x+1-12log x-1-4x-1+C'=12log x+1x-1-4x-1+C'

Page No 19.177:

Question 44:

x3-1x3+x dx

Answer:

We have,I= x3-1dxx3+x
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
               1x3+x  x3-1              x3+x         -   -                 -x-1 x3-1x3+x=1-x+1x3+xx3-1x3+x=1-x+1xx2+1     .....1Let x+1xx2+1=Ax+Bx+Cx2+1x+1xx2+1=Ax2+1+Bx+Cxxx2+1x+1=Ax2+A+Bx2+Cxx+1=A+Bx2+Cx+A
Equating coefficient of like terms
A + B = 0
C = 1
A = 1
B = –1
x+1xx2+1=1x+-x+1x2+1          .....2Using 1 & 2 x3-1dxx3+x= 1-1x+xx2+1-1x2+1dx=dx-dxx+x dxx2+1-dxx2+1 Putting x2+1=t2x dx=dtx dx=dt2I= dx-dxx+12dtt-dxx2+1=x-log x+12log t-tan-1x+C=x-log x+12log x2+1-tan-1x+C

Page No 19.177:

Question 45:

x2+x+1x+12 x+2 dx

Answer:

We have,I=x2+x+1x+12 x+2dxLet x2+x+1x+12 x+2=Ax+1+Bx+12+Cx+2x2+x+1x+12 x+2=Ax+1 x+2+Bx+2+Cx+12x+12 x+2x2+x+1=Ax2+x+2x+2+Bx+2B+Cx2+2x+1x2+x+1=A+Cx2+3A+B+2Cx+2A+2B+CEquating coefficient of like terms.A+C=1                 .....13A+B+2C=1      .....22A+2B+C=1      .....3Solving these three equations we getA=-2B=1C=3Hence, x2+x+1x+12 x+2=-2x+1+1x+12+3x+2I=-2dxx+1+dx+12+3dxx+2=-2 log x+1-1x+1+3 log x+2+C

Page No 19.177:

Question 46:

1x x4+1 dx

Answer:

We have,I= dxxx4+1=x3 dxx4 x4+1Putting x4=t4x3 dx=dtx3 dx=dt4I=14dttt+1Let 1tt+1=At+Bt+11tt+1=At+1+Bttt+11=At+1+BtPutting t+1=0t=-11=A×0+B-1B=-1Putting t=01=A1+B×0A=1 I=14dtt-14dtt+1=14log t-14log t+1+C=14log tt+1+C=14log x4x4+1+C

Page No 19.177:

Question 47:

Evaluate the following integrals:

1xx3+8dx

Answer:

Let I=1xx3+8dxWe express1xx3+8==Ax+Bx2+Cx+Dx3+81=Ax3+8+Bx2+Cx+DxEquating the coefficients of x3, x2, x and constants, we get0=A+B      and      0=C      and      0=D      and      1=8Aor A=18      and      B=-18      and      C=0      and      D=0I=18x+-18x2x3+8dx     =181xdx-1243x2x3+8 dx     =18logx-124logx3+8+cHence, 1xx3+8dx=18logx-124logx3+8+c

Page No 19.177:

Question 48:

31-x 1+x2 dx

Answer:

We have,I= 3 dx1-x 1+x2=3dx1-x 1+x2Let 11-x 1+x2=A1-x+Bx+Cx2+111-x x2+1=Ax2+1+Bx+C 1-x1-x x2+11=Ax2+A+Bx-Bx2+C-Cx1=A-Bx2+B-Cx+A+CEquating coefficients of like terms.A-B=0     .....1B-C=0     .....2A+C=1     .....3Solving 1, 2 and 3, we getA=12, B=12, C=1211-x x2+1=121-x+x2+12x2+1 3 dx1-x x2+1=32dx1-x+32x dxx2+1+32dxx2+1Putting x2+1=tx dx=dt2I=32dx1-x+34dtt+32dxx2+1=32log  1-x-1+34log t+32×tan-1x+C=-32log 1-x+34log 1+x2+32tan-1x+C=-34×2 log 1-x+34log 1+x2+342 tan-1 x+C=34log 1+x2-log 1-x2+342 tan-1 x+C=34log 1+x21-x2+2 tan-1 x+C

Page No 19.177:

Question 49:

cos x1-sin x3 2+sin x dx

Answer:

We have, I=cos x1-sin x3 2+sin x dxLet, sinx=tcosx dx=dtNow, integration becomes,I=dt1-t3 2+t   =-dtt-13 t+2 Let, 1t-13 t+2=At-1+Bt-12+Ct-13+D t+2     .....11=At-12t+2+Bt-1t+2+Ct+2+Dt-13            .....2

Putting t=1 in 2, we get1=3CC=13Putting t=-2 in 2, we get1=D-2-131=-27DD=-127Putting t=0 in 2, we get1=2A-2B+2C-D1=2A-2B+23+1272A-2B=827A-B=427Putting t=2 in 2, we get1=4A+4B+4C+D1=4A+4B+43-127A+B=-227Now, A-B=427 and A+B=-227 A=127 and B=-19
Substituting the values of A, B, C and D in 1, we get1t-13 t+2=127t-1-19t-12+13t-13+-1 27t+2Now, integration becomes I=-127t-1-19t-12+13t-13+-1 27t+2dt  =-127log t-1+19t-1-16t-12-127log t+2+C  =-127log sin x-1-19sin x-1+16sin x-12+127log sin x+2+C  =-127log 1-sin x+191-sin x+161-sin x2+127log 2+sin x+C

Page No 19.177:

Question 50:

Evaluate the following integrals:

2x2+1x2x2+4dx

Answer:

Let I=2x2+1x2x2+4dxWe express2x2+1x2x2+4=Ax2+Bx2+42x2+1=Ax2+4+Bx2Equating the coefficients of x2 and constants, we get2=A+B      and      1=4Aor A=14      and      B=74I=14x2+74x2+4dx     =141x2dx+741x2+4 dx     =-14x+78tan-1x2+cHence, 2x2+1x2x2+4dx=-14x+78tan-1x2+c

Page No 19.177:

Question 51:

cos x1-sin x 2-sin x dx

Answer:

We have,I= cos x dx1-sin x 2-sin xPutting sin x=tcos x dx=dtI=dt1-t 2-t=dtt-1 t-2Let  1t-1 t-2=At-1+Bt-21t-1 t-2=At-2+Bt-1t-1 t-21=At-2+Bt-1Putting t-1=0t=11=A1-2+B×0A=-1Putting t-2=0t=21=A×0+B2-1B=1I=-dtt-1+dtt-2=-log t-1+log t-2+C=logt-2t-1+C=log sin x-2sin x-1+C=log 2-sin x1-sin x+C

Page No 19.177:

Question 52:

2x+1x-2 x-3 dx

Answer:

We have,I= 2x+1dxx-2 x-3Let 2x+1x-2 x-3=Ax-2+Bx-32x+1x-2 x-3=Ax-3+Bx-2x-2 x-32x+1=Ax-3+Bx-2Putting x-3=0x=37=A×0+B×3-2B=7Putting x-2=0x=25=A-1A=-5I=-5dxx-2+7dxx-3=-5 log x-2+7 log x-3+C=log x-37-logx-25+C=log x-37x-25+C

Page No 19.177:

Question 53:

1x2+1 x2+2 dx

Answer:

We have,I= dxx2+1 x2+2Putting x2=tThen,1x2+1 x2+2=1t+1 t+2Let 1t+1 t+2=At+1+Bt+21t+1 t+2=At+2+Bt+1t+1 t+21=At+2+Bt+1Putting t+2=0t=-21=A×0+B-1B=-1Putting t+1=0t=-11=A-1+2+B×0A=11t+1 t+2=1t+1-1t+21x2+1 x2+2=1x2+1-1x2+22I= dxx2+12-dxx2+22=tan-1 x-12tan-1 x2+C

Page No 19.177:

Question 54:

1x x4-1 dx

Answer:

We have,I= dxxx4-1= x3 dxx4 x4-1Putting x4=t4x3 dx=dtx3 dx=dt4I=14dttt-1Let 1tt-1=At+Bt-11tt-1=At-1+B ttt-11=At-1+BtPutting t-1=0t=11=A×0+B1B=1Putting t=01=A0-1+B×0A=-1I=-14dtt+14dtt-1=-14log t+14log t-1+C=14log t-1t+C=14log x4-1x4+C

Page No 19.177:

Question 55:

1x4-1 dx

Answer:

We have,I=dxx4-1=dxx2-1 x2+1=dxx-1 x+1 x2+1Let 1x-1 x+1 x2+1=Ax-1+Bx+1+Cx+Dx2+11x-1 x+1 x2+1=Ax2+1 x+1+Bx-1 x2+1 Cx+D x-1 x+1x-1 x+1 x2+11=Ax2+1 x+1+B x-1 x2+1+Cx+D x2-11=Ax3+x2+x+1+Bx3+x-x2-1+Cx3-Cx+Dx2-D1=A+B+C x3+x2A-B+D+xA+B-C+A-B-DEquating the coefficients of like terms.A+B+C=0               .....1A-B+D=0               .....2A+B-C=0              .....3A-B-D=1              .....4Solving these four equations we getA=14, B=-14, C=0, D=-121x-1 x+1 x2+1=14x-1-14x+1-12x2+1I=14 dxx-1-14dxx+1-12dxx2+1=14log  x-1 -14log x+1-12tan-1 x+C'=14log x-1x+1-12tan-1 x+C'

Page No 19.177:

Question 56:

Find 2xx2+1x2+22dx

Answer:

 2xx2+1x2+22dx

Let x2=y2xdx=dydx=dy2x

 2xx2+1x2+22dx=dyy+1y+22Let 1y+1y+22=Ay+1+By+2+Cy+22      .....11=Ay+22+By+1y+2+Cy+1       .....2Putting y=-2 in 21=C-2+1C=-1

Putting y=-1 in 21=A-1+221=A1A=1

Putting y=0 in 21=4A+B2+C1=4+2B-11=3+2B-2=2BB=-1

Substituting the values of A, B and C in (1)

1y+1y+22=1y+1-1y+2-1y+22dyy+1y+22=dyy+1-dyy+2-dyy+22=logy+1-logy+2+1y+2+C
Hence, 2xx2+1x2+22dx = =logx2+1-logx2+2+1x2+2+C

Page No 19.177:

Question 57:

Evaluate the following integrals:
x2(x-1) (x2+1)dx

Answer:

Ix2(x-1) (x2+1)dx

x2(x-1)(x2+1)=A(x-1)+Bx+Cx2+1                           = Ax2+1+Bx+Cx-1(x-1)(x2 + 1)x2(x-1)(x2+1)=(A+B)x2+(C-B)x+A-C(x-1)(x2+1)

Comparing coefficients, we get

A+B=1; C-B=0 and A-C=0Solving these equations, we getA=B=C=12 I=121(x-1)dx+12xx2+1dx+121x2+1dx         =12lnx-1+14lnx2+1+12tan-1x+c

Page No 19.177:

Question 58:

Evaluate the following integrals:

x2x2+a2x2+b2dx

Answer:

Let I=x2x2+a2x2+b2dxWe expressx2x2+a2x2+b2=Ax2+a2+Bx2+b2x2=Ax2+b2+Bx2+a2Equating the coefficients of x2 and constants, we get1=A+B      and      0=b2A+a2Bor A=-a2b2-a2      and      B=b2b2-a2I=-a2b2-a2x2+a2+b2b2-a2x2+b2dx     =-a2b2-a21x2+a2dx+b2b2-a21x2+b2 dx     =-ab2-a2tan-1xa+bb2-a2tan-1xb+cHence, x2x2+a2x2+b2dx=-ab2-a2tan-1xa+bb2-a2tan-1xb+c

Page No 19.177:

Question 59:

1cos x 5-4 sin x dx

Answer:

We have,I=dxcos x 5-4 sin x=cos x dxcos2x 5-4 sin x=cos x dx1-sin2x 5-4 sin x=cos x dx1-sin x 1+sin x 5-4 sin xPutting sin x=tcos x dx=dtI=dt1-t 1+t 5-4tLet 11-t 1+t 5-4t=A1-t+B1+t+C5-4t11-t 1+t 5-4t=A1+t 5-4t+B1-t 5-4t+C1-t 1+t1-t 1+t 5-4t1=A1+t 5-4t+B1-t 5-4t+C1-t 1+tPutting 1+t=0t=-11=B2 5+4B=118Putting  1-t=0t=11=A 2 5-4+B×0+C×0A=12Putting 5-4t=04t=5t=541=C 1-54 1+541=C -14 94C=-169I=12dt1-t+118dt1+t-169dt5-4t=12 log 1-t-1+118 log 1+t-169× log 5-4t-4+C=118 log 1+t-12 log 1-t+49log 5-4t+C=118 log 1+sin x-12 log 1-sin x+49 log 5-4 sin x+C



Page No 19.178:

Question 60:

1sin x 3+2 cos x dx

Answer:

We have,I=dxsin x 3+2 cos x=sin x dxsin2x 3+2 cos x=sin x dx1-cos2x 3+2 cos x=sin x dx1-cos x 1+cos x 3+2 cos xPutting cos x=t-sin x dx=dtsin x dx=-dtI=-dt1-t 1+t 3+2t=dtt-1 t+1 3+2tLet 1t-1 t+1 3+2t=At-1+Bt+1+C3+2t1t-1 t+1 3+2t=A t+1 3+2t+B t-1 3+2t+C t+1 t-1t-1 t+1 3+2t1=A t+1 3+2t+B t-1 3+2t+C t+1 t-1Putting t+1=0t=-11=A×0+B -2 3-2+C×01=B -2B=-12Putting  t-1=0t=11=A 2 5+B×0+C×0A=110Putting  3+2t=0t=-321=A×0+B×0+C -32+1 -32-11=C -12 -52C =45Then,I=110dtt-1-12dtt+1+45dt3+2t=110 log t-1-12 log t+1+45× log 3+2t2+C=110 log t-1-12 log t+1+25log 3+2t+C=110 log cos x-1-12 log cos x+1+25 log 3+2 cos x+C

Page No 19.178:

Question 61:

1sin x+sin 2x dx

Answer:

We have,I=dxsin x +sin 2x =dxsin x+2 sin x cos x=dxsin x 1+2 cos x=sin x dxsin2x 1+2 cos x=sin x dx1-cos2x 1+2 cos x=sin x dx1-cos x 1+cos x 1+2 cos xPutting cos x=t-sin x dx=dtsin x dx=-dtI=-dt1-t 1+t 1+2t=dtt-1 t+1 1+2tLet 1t-1 t+1 1+2t=At-1+Bt+1+C1+2t1t-1 t+1 1+2t=A t+1 1+2t+B t-1 1+2t+C t-1 t+1t-1 t+1 1+2t1=A t+1 1+2t+B t-1 1+2t+C t-1 t+1Putting t+1=0t=-11=B -1-1 1-21=B -2 -1B=12Putting t-1=0t=11=A 1+1 1+21=A23A=16Putting 1+2t=0t=-121=A×0+B×0+C -12-1 -12+11=C -32 12C=-43Then,I=16dtt-1+12dtt+1-43dt1+2t=16 log t-1+12 log t+1-43×log 1+2t2+C=16 log t-1+12 log t+1-23log 1+2t+C=16 log cos x-1+12 log cos x+1-23 log 1+2 cos x+C

Page No 19.178:

Question 62:

x+1x 1+xex dx

Answer:

We have,I=x+1x 1+xex dxI=exx+1exx 1+xex dxPut ex=texx+1dx=dtI=dtt 1+t         .....1Let,1t 1+t=At +B1+t1=At+1+Bt         .....2Putting t=0 in 2, we obtain A=1Putting t=-1 in 2, we obtain B=-1I=1t -11+t dtI=logt-logt+1+CI=logtt+1 +CI=logxexxex+1 +C

Page No 19.178:

Question 63:

x2+1 x2+2x2+3 x2+4 dx

Answer:

We have,I= x2+1 x2+2x2+3 x2+4Putting x2=tThen,x2+1 x2+2x2+3 x2+4=t+1 t+2t+3 t+4=t2+3t+2t2+7t+12
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
                       1t2+7t+12   t2+3t+2                        t2+7t+12                  -   -    -                         -4t-10
t2+3t+2t2+7t+12=1-4t+10t2+7t+12t2+3t+2t2+7t+12=1-4t+10t+3 t+4     .....1Let 4t+10t+3 t+4=At+3+Bt+44t+10t+3 t+4=At+4+Bt+3t+3 t+44t+10=At+4+Bt+3Putting t+4=0t=-4-16+10=B-1B=6Putting t+3=0t=-3-12+10=A-3+4A=-24t+10t+3 t+4=-2t+3+6t+4       .....2From 1 & 2t2+3t+2t2+7t+12=1+2t+3-6t+4x2+1 x2+2dxx2+3 x2+4=dx+2dxx2+32-6dxx2+22                                 =x+23tan-1 x3-62tan-1 x2+C                                =x+23 tan-1 x3-3 tan-1 x2+C

Page No 19.178:

Question 64:

4x4+3x2+2 x2+3 x2+4 dx

Answer:

We have,I= 4x4+3dxx2+2 x2+3 x2+4Putting x2=tThen,4x4+3x2+2 x2+3 x2+4=4t2+3t+2 t+3 t+4Let 4t2+3t+2 t+3 t+4=At+2+Bt+3+Ct+44t2+3t+2 t+3 t+4=At+3 t+4+Bt+2 t+4+Ct+2 t+3t+2 t+3 t+44t2+3=At+3 t+4+Bt+2 t+4+Ct+2 t+3Putting t+3=0t=-34×-32+3=B-3+2 -3+439=B-1B=-39Putting t+2=0t=-24-22+3=A-2+3 -2+419=A×1×2A=192Let t+4=0t=-44×-42+3=C-4+2 -4+367=C-2 -1C=6724t2+3t+2 t+3 t+4=192t+2-39t+3+672t+44x4+3x2+2 x2+3 x2+4=192x2+2-39x2+3+672x2+4I=192dxx2+22-39dxx2+32-672dxx2+22=192×12tan-1 x2-393tan-1 x3-672×12tan-1 x2+C=1922 tan-1 x2-393tan-1 x3-674tan-1 x2+C

Page No 19.178:

Question 65:

x4x-1 x2+1 dx

Answer:

We have,I= x4 dxx-1 x2+1= x4-1+1x-1 x2+1dx= x4-1dxx-1 x2+1+dxx-1 x2+1=x2-1 x2+1 dxx-1 x2+1+dxx-1 x2+1= x-1 x+1dxx-1+dxx-1 x2+1=x+1dx+dxx-1 x2+1      .....1Let 1x-1 x2+1=Ax-1+Bx+Cx2+11x-1 x2+1=Ax2+1+Bx+C x-1x-1 x2+11=Ax2+A+Bx2-Bx+Cx-C1=A+Bx2+C-Bx+A-CEquating coefficients of like termsA+B=0     .....1C-B=0     .....2A-C=1     .....3Solving 1, 2 and 3, we getB=-12, A=12, C=-121x-1 x2+1=12x-1+-x2-12x2+11x-1 x2+1=12x-1-12xx2+1-12x2+1     .....2From 1 & 2I=x+1dx+12dxx-1-12x dxx2+1-12dxx2+1Putting x2+1=t2x dx=dtx dx=dt2I=x+1dx+12dxx-1-14dtt-12dxx2+1=x22+x+12log x-1-14log t-12tan-1x+C=x22+x+12log x-1-14log x2+1-12tan-1x+C

Page No 19.178:

Question 66:

Evaluate the following integrals:

x2x4-x2-12dx

Answer:

Let I=x2x4-x2-12dxWe expressx2x4-x2-12=x2x4-4x2+3x2-12                   =x2x2-4x2+3                   =Ax2-4+Bx2+3x2=Ax2+3+Bx2-4Equating the coefficients of x2 and constants, we get1=A+B      and      0=3A-4Bor A=47      and      B=37I=47x2-4+37x2+3dx     =471x2-4dx+371x2+3 dx     =47×14logx-2x+2+37tan-1x3+c     =17logx-2x+2+37tan-1x3+cHence, x2x4-x2-12dx=17logx-2x+2+37tan-1x3+c

Page No 19.178:

Question 67:

Evaluate the following integrals:

x21-x4dx
 

Answer:

Let I=x21-x4dxWe expressx21-x4=x21-x21+x2           =A1-x2+B1+x2x2=A1+x2+B1-x2Equating the coefficients of x2 and constants, we get1=A-B      and      0=A+Bor A=12      and      B=-12I=121-x2+-121+x2dx     =1211-x2dx-1211+x2 dx     =12×12log1+x1-x-12tan-1x+c     =14log1+x1-x-12tan-1x+cHence, x21-x4dx=14log1+x1-x-12tan-1x+c

Page No 19.178:

Question 68:

Evaluate the following integrals:

x2x4+x2-2dx​

Answer:

Let I=x2x4+x2-2dxWe expressx2x4+x2-2=x2x4+2x2-x2-2                 =x2x2+2x2-1                 =Ax2+2+Bx2-1x2=Ax2-1+Bx2+2Equating the coefficients of x2 and constants, we get1=A+B      and      0=-A+2Bor A=23      and      B=13I=23x2+2+13x2-1dx     =231x2+2dx+131x2-1 dx     =23tan-1x2+16logx-1x+1+cHence, x2x4+x2-2dx=23tan-1x2+16logx-1x+1+c

Page No 19.178:

Question 69:

(x2+1) (x2+4)(x2+3) (x2-5) dx

Answer:

I(x2+1) (x2+4)(x2+3) (x2-5) dx

Since,

x2+1x2+4x2+3x2-5=x2+3-2x2-5+9x2+3x2-5x2+1x2+4x2+3x2-5=x2+3x2-5+9x2+3-2x2-5-18x2+3x2-5

x2+1x2+4x2+3x2-5=1+9x2-5-2x2+3-18x2+3x2-5         ...(i)


Let I1=1(x2+3)(x2-5) and x2 = y

1y+3y-5 =Ay+3+By-5                               =Ay-5+By+3y+3y-51y+3y-5 =A+By-5A+3By+3y-5

Comparing coefficients, we get

A+B=0 and 5A+3B=-1By solving the equations, we getA=-18 and B=18

From (i), we get

I=1+9x2-5-2x2+3-18-18x2+3+18x2-5dx

I=1+274x2-5+1x2+3dxI=1dx+274x2-5dx+1x2+3dx I=x+2785lnx-5x+5+143tan-1x3+c

Page No 19.178:

Question 70:

2 cos x1-sin x 1+sin2xdx

Answer:


Let sin x = t  cosx dx = dt2dt1-t1+t2Using partial fraction21-t1+t2 = A1-t+ Bt+C1+t2A+At2+Bt+C-Bt2-Ct=2A+C=2A=2-C    .....1Bt-Ct=0       .....2And At2-Bt2=0    .....3Substitute 1 in 3 B=1Put B=1 in 2C=1Put C=1 in 1Hence, A =1, B = 1, C = 12dt1-t1+t2 = dt1-t+1+t1+t2dt=dt1-t + dt1+t2+t dt1+t2= -ln (1-t) + tan-1t + 12ln1+t2= ln 1+t21-t + tan-1t + CReplacing the value of t=ln 1+sin2x1-sinx + tan-1(sin x) + C



Page No 19.190:

Question 1:

x2+1x4+x2+1 dx

Answer:

We have,I= x2+1dxx4+x2+1Dividing numerator and denominator by x2, we getI= 1+1x2dxx2+1+1x2= 1+1x2dxx2+1x2-2+3= 1+1x2dxx-1x2+3Putting x-1x=t1+1x2dx=dtI= dtt2+3=dtt2+32=13tan-1 t3+C=13tan-1 x-1x3+C=13 tan-1 x2-13 x+C

Page No 19.190:

Question 2:

cot θd θ

Answer:

We have,I=cot θ dθPutting cot θ=t2-cosec2 θ dθ=2t dtdθ=-2t dtcosec2 θdθ=-2t dt1+cot2 θdθ=-2t dt1+t4I= t-2t dt1+t4=-2t21+t4dt=-t2+1+t2-1t4+1dt=-t2+1t4+1dt-t2-1dtt4+1Dividing numerator and denominator by t2I=-1+1t2t2+1t2dt-1-1t2t2+1t2dt=-1+1t2dtt2+1t2-2+2-1-1t2dtt2+1t2+2-2=-1+1t2dtt-1t2+22-1-1t2dtt+1t2-22Putting t-1t=p1+1t2dt=dpPutting t+1t=q1-1t2dt=dqI=- dpp2+22-dqq2-22=-12tan-1 p2-122log q-2q+2+C=-12tan-1 t-1t2-122log t+1t-21+1t+2+C=-12 tan-1 t2-12 t-122log t2+1-2tt2+1+2t+C=-12tan-1 cot θ-12cot θ-122log cot θ+1-2 cot θcot θ+1+2 cot θ+C

Page No 19.190:

Question 3:

x2+9x4+81 dx

Answer:

We have,I= x2+9x4+81dxDividing numerator and denominator by x2I=1+9x2dxx2+81x2=1+9x2dxx2+9x2-2×x×9x+2×x×9x=1+9x2dxx-9x2+182Putting x-9x=t1+9x2dx=dtI=dtt2+182=dtt2+322=132tan-1 t32+C=132tan-1 x-9x32+C=132tan-1 x2-932x+C

Page No 19.190:

Question 4:

1x4+x2+1 dx

Answer:

We have,I= dxx4+x2+1=12 2 dxx4+x2+112x2+1-x2-1x4+x2+1dx12x2+1x4+x2+1dx-12x2-1x4+x2+1dxDividing numerator and denominator by x2I=121+1x2x2+1+1x2dx-121-1x2x2+1+1x2dx=121+1x2x2+1x2-2+3dx-121-1x2dxx2+1x2+2-1=121+1x2dxx-1x2+32-121-1x2dxx+1x2-12Putting x-1x=t1+1x2dx=dtPutting x+1x=p1-1x2dx=dpI=12dtt2+32-12dpp2-12=12×13tan-1 t3-12×12×1log p-1p+1+C=123tan-1 x-1x3-14log x+1x-1x+1x+1+C=123tan-1 x2-1x3-14log x2-x+1x2+x+1+C

Page No 19.190:

Question 5:

x2-3x+1x4+x2+1 dx

Answer:

We have,I=x2-3x+1x4+x2+1dx=x2+1dxx4+x2+1-3x dxx4+x2+1                .....1=I1-3I2 where I1=x2+1dxx4+x2+1, I2=x dxx4+x2+1I1=x2+1x4+x2+1dxDividing numerator & denominator by x2I1=1+1x2dxx2+1x2+1=1+1x2dxx2+1x2-2+3=1+1x2dxx-1x2+32Let x-1x=t1+1x2dx=dtI1=dtt2+32I1=13tan-1 t3+C1I1=13tan-1 x-1x3+C1              .....2I2=x dxx4+x2+1Putting x2=t2x dx=dtx dx=dt2I2=12dtt2+t+1=12dtt2+t+14+34=12dtt+122+322=132×12tan-1 t+1232+C2=13tan-12t+13+C2I2=13tan-1 2x2+13+C2        ...3From equating 1, 2 and 3 we haveI=13tan-1 x-1x3+C1-3×13tan-1 2x2+13+C2=13tan-1 x2-13x-3tan-1 2x2+13+C where C=C1+3C2

Page No 19.190:

Question 6:

x2+1x4-x2+1 dx

Answer:

We have,I=x2+1x4-x2+1dxDividing numerator and denominator by x2I=1+1x2x2+1x2-1dx=1+1x2dxx2+1x2-2+1=1+1x2dxx-1x2+1Putting x-1x=t1+1x2dx=dtI=dtt2+12=tan-1t+C=tan-1x-1x+C=tan-1 x2-1x+C

Page No 19.190:

Question 7:

x2-1x4+1 dx

Answer:

We have,I= x2-1x4+1dxDividing numerator and denominator by x2=1-1x2x2+1x2dx=1-1x2dxx2+1x2+2-2=1-1x2dxx+1x2-22Putting x+1x=t1-1x2dx=dtI=dtt2-22=122log t-2t+2+C=122log x+1x-2x+1x+2+C=122log x2-2x+1x2+2x+1+C

Page No 19.190:

Question 8:

x2+1x4+7x2+1 dx

Answer:

We have,I= x2+1x4+7x2+1dxDividing numerator and denominator by x2I=1+1x2x2+7+1x2dx=1+1x2dxx2+1x2-2+91+1x2dxx-1x2+32Putting x-1x=t1+1x2dx=dtI=dtt2+32=13tan-1 t3+C=13tan-1 x-1x3+C=13tan-1 x2-13x+C

Page No 19.190:

Question 9:

x-12x4+x2+1 dx

Answer:

We have,I=x-12dxx4+x2+1=x2-2x+1x4+x2+1dx=x2+1x4+x2+1dx-2x dxx4+x2+1=I1-I2 where,I1=x2+1dxx4+x2+1I2= 2x dxx4+x2+1Now,I1= x2+1x4+x2+1dxDividing numerator and denominator by x2I1=1+1x2x2+1x2+1dxI1=1+1x2dxx2+1x2-2+3I1=1+1x2dxx-1x2+32Putting x-1x=t1+1x2dx=dtI1= dtt2+32=13tan-1 t3+C1=13tan-1 x-1x3+C1=13tan-1 x2-13x+C1AndI2=2x dxx4+x2+1Putting x2=t2x dx=dtI2= dtt2+t+1=dtt2+t+122-122+1=dtt+122+322=23tan-1 t+1232+C2=23tan-1 2t+13+C2I=13tan-1 x2-13x-23tan-1 2x2+13+C

Page No 19.190:

Question 10:

1x4+3x2+1 dx

Answer:

We have,I= dxx4+3x2+1=12 2 dxx4+3x2+1=12x2+1-x2-1x4+3x2+1dx=12x2+1x4+3x2+1dx-12x2-1x4+3x2+1dxDividing numerator and denominator by x2=121+1x2x2+1x2+3dx-121-1x2dxx2+1x2+3=121+1x2x2+1x2-2+5dx-121-1x2dxx2+1x2+2+1=121+1x2dxx-1x2+52-121-1x2dxx+1x2+12Putting x-1x=t1+1x2dx=dtPutting x+1x=p1-1x2dx=dpI=12dtt2+52-12dpp2+12=125tan-1 t5-12tan-1 p+C=125tan-1 x-1x5-12tan-1 x+1x+C=125tan-1 x2-15x-12tan-1 x2+1x+C

Page No 19.190:

Question 11:

Evaluate the following integrals:

1sin4x+sin2x cos2x+cos4xdx

Answer:

Let I=1sin4x+sin2x cos2x+cos4xdx       =1sin2x+cos2x2-sin2x cos2xdx       =11-sin2x cos2xdx       =1cos4x1cos4x-sin2xcos2xdx       =sec2x1+tan2xsec4x-tan2xdx       Let tanx=t       On differentiating both sides, we get       sec2x dx=dtI=1+t21+t22-t2dt     =1+t2t4+t2+1dt     =1t2+1t2+1+1t2dt     =1t2+1t-1t2+3dt       Let t-1t=u       On differentiating both sides, we get       1+1t2 dt=duI=1u2+3du     =13tan-1u3+c     =13tan-1t-1t3+c     =13tan-1tanx-cotx3+cHence, 1sin4x+sin2x cos2x+cos4xdx=13tan-1tanx-cotx3+c



Page No 19.196:

Question 1:

1x-1 x+2 dx

Answer:

We have,I= dxx-1 x+2Putting x+2=t2dx=2t dtI=2t dtt2-2-1t= 2 dtt2-32=2×123log t-3t+3+C=13log x+2-3x+2+3+C

Page No 19.196:

Question 2:

1x-1 2x+3 dx

Answer:

We have,I= dxx-1 2x+3Putting 2x+3=t2x=t2-32Diff both sidesdx=t dtI= t dtt2-32-1t=2 dtt2-3-2=2dtt2-5=2dtt2-52=2×125log t-5t+5+C=15log 2x+3-52x+3+5+C

Page No 19.196:

Question 3:

x+1x-1 x+2 dx

Answer:

We have,I= x+1x-1 x+2dxPutting x+2=t2x=t2-2Diff both sidesdx=2t dtI= t2-2+12t dtt2-2-1t=2 t2-1t2-3dt=2t2-3+2t2-3dt=2 t2-3t2-3dt+4dtt2-3=2dt+4dtt2-32=2t+4×123log t-3t+3+C=2x+2+23log x+2-3x+2+3+C

Page No 19.196:

Question 4:

x2x-1 x+2 dx 

Answer:

We have,I= x2x-1 x+2dxPutting x+2=t2x=t2-2Diff both sidesdx=2t dtI= t2-22t2-2-1t2 t dt=2 t2-22dtt2-3=2 t4-4t2+4t2-3dtDividing numerator by denominator, we get            t2-1t2-3 t4-4t2+4           t4-3t2         -  +               -t2+4                -t2+3               +   -                    1       I=2t2-1+1t2-3dt                   =2 t2 dt-2dt+2dtt2-32=2t33-2t+2×123log t-3t+3+C=23x+23-2x+2+13log x+2-3x+2+3+C=23x+232-2x+2+13log x+2-3x+2+3+C

Page No 19.196:

Question 5:

xx-3 x+1 dx

Answer:

We have,I= x dxx-3 x+1Putting x+1=t2x=t2-1Diff both sidesdx=2t dtI= t2-12t dtt2-1-3t=2 t2-1t2-4dt=2t2-4+3t2-4dt=2t2-4t2-4dt+6 dtt2-22=2 dt+6dtt2-22=2t+6×12×2log t-2t+2+C=2x+1+32log t-2t+2+C=2x+1+32log x+1-2x+1+2+C

Page No 19.196:

Question 6:

1x2+1 x dx

Answer:

We have,I= dxx2+1 xPutting x=t2dx=2t dtI= 2t dtt22+1t=2 dtt4+1= t2+1-t2-1t4+1dt=t2+1t4+1dt-t2-1t4+1dtDividing numerator & denominator by t2I=1+1t2t2+1t2dt- 1-1t2dtt2+1t2= 1+1t2dtt2+1t2-2+2- 1-1t2dtt2+1t2+2-2= 1+1t2dtt-1t2+22- 1-1t2dtt+1t2-22Putting t-1t=p1+1t2dt=dpPutting t+1t=q1-1t2dt=dqI=dpp2+22-dqq2-22=12tan-1 p2-122log q-2q+2+C=12tan-1 t-1t2-122log t+1t-2t+1t+2+C=12tan-1 t2-12t-122log  t2-2t+1t2+2t+1+C=12tan-1 x-12x-122log x-2x+1x+2x+1+C

Page No 19.196:

Question 7:

xx2+2x+2 x+1 dx

Answer:

We have,I= x dxx2+2x+2 x+1= x dxx+12+1 x+1Putting x+1=t2x=t2-1Diff both sidesdx = 2t dtI= t2-12t dtt22+1 t=2 t2-1dtt4+1Dividing numerator and denominator by t2I=21-1t2t2+1t2dt

=21-1t2dtt2+1t2+2-2=2 1-1t2dtt+1t2-22Putting t+1t=p1-1t2dt=dpI=2 dpp2-22=2×122log p-2p+2+C=12log p-2P+2+C=12log t+1t-2t+1t+2+C=12log  t2-2t+1t2+2t+1+C=12log x+1-2x+1+1x+1+2x+1+1+C=12log x+2-2x+1x+2+2x+1+C

Page No 19.196:

Question 8:

1x-1 x2+1 dx

Answer:

We have,I= dxx-1 x2+1Putting x-1=1tdx=-1t2dtI=-1t2dt1t 1+1t2+1= -1tdt1+1t2+2t+1= -1tdtt2+1+2t+t2t= -dt2t2+2t+1=-12  dtt2+t+12=-12 dtt2+t+14-14+12=-12  dtt+122+122=-12log t+12+t+122+14+C where t=1x-1

Page No 19.196:

Question 9:

1x+1 x2+x+1 dx

Answer:

We have,I= dxx+1 x2+x+1Putting x+1=1tdx=-1t2dtI= -1t2dt1t1t-12+-1+11t= -1t2dt1t1t2-+1+2t1t= -1tdtt2+t-2t+1t=- dtt2-t+1=-dtt2-t+14-14+1=-dtt-122+322=-log t-12+t-122+322+C=-log  t-12+t2-t+1+C=-log 1x+1-12+1x+12-1x+1+1+C=-log 1x+1-12+x+12-x+1+1x+1+C=-log 1x+1-12+x2+x+1x+1+C

Page No 19.196:

Question 10:

1x2-1 x2+1 dx

Answer:

We have,I= dxx2-1 x2+1Putting x=1tdx=-1t2dtI= -1t2dt1t2-1 1t2+1= -1t2 dt1-t2t2×1+t2t= -t dt1-t2 1+t2Putting 1+t2=u2t2=u2-12t dt=2u dut dt=u duI=-u du1-u2+1u=- du2-u2=- du22-u2=-122log u+2u-2+C=-122log 1+t2+21+t2-2+C=-122log 1+1x2+21+1x2-2+C=-122log x2+1+2xx2+1-2x+C

Page No 19.196:

Question 11:

xx2+4 x2+1 dx

Answer:

We have,I= x dxx2+4 x2+1Putting x2=t 2x dx=dtx dx=dt2I=12 dtt+4 t+1Again Putting  t+1=p2t=p2-1dt=2p dpI=12 2p dpp2-1+4p= dpp2+3=dpp2+32=13tan-1 p3+C=13tan-1 t+13+C=13 tan-1 x2+13+C

Page No 19.196:

Question 12:

11+x2 1-x2 dx

Answer:

We have,I= dx1+x2 1-x2Putting x=1tdx=-1t2dtI= -1t2dt1+1t2 1-1t2= -1t2dtt2+1t2 t2-1t=- t dtt2+1 t2-1Again Putting t2-1=u22t dt=2u dut dt=u duI=- u duu2+2u=- duu2+22=-12tan-1 u2+C=-12 tan-1 t2-12+C=-12tan-1 1x2-12+C=-12tan-1 1-x22x2+C

Page No 19.196:

Question 13:

12x2+3 x2-4 dx

Answer:

​We have,I=dx2x2+3 x2-4Putting x=1tdx=-1t2dtI=-1t2dt2t2+3 1t2-4=-1t2 dt2+3t2t2×1-4t2t=-t dt2+3t2 1-4t2Again Putting 1-4t2=u2-8t dt=2u dut dt=-u4 duI=14u du2+3 1-u24 u=144 du8+3-3u2=du11-3u2=13du113-u2=13du1132-u2=13×12×113 log 113+u113-u+C=1233 log 11+3 u11-3 u+C=1233 log 11+3 1-4t211-3 1-4t2+C=1233 log 11+3-12t211-3-12t2+C=1233 log 11+3-12x211-3-12x2+C=1233 log 11x+3x2-1211x-3x2-12+C

Page No 19.196:

Question 14:

xx2+4 x2+9 dx

Answer:

We have,I=x dxx2+4 x2+9Putting x2=t2x dx= dtx dx=dt2I=12dtt+4 t+9Again Putting t+9=u2dt=2u duI=122u duu2-9+4 u=duu2-5=duu2-52=125 log u-5u+5+C=125 log t+9-5t+9+5+C=125 log x2+9- 5x2+9+ 5+C



Page No 19.197:

Question 1:

Write a value of 1+cot xx+log sin x dx.

Answer:

Let I=1+cot xx+log sin xdxLet x +log sin x=t1+1sin x×cos x dx=dt1+cot xdx=dtI=dtt     =log t+C     =log x+log sin x+C

Page No 19.197:

Question 2:

Write a value of e3 log x x4 dx.

Answer:

e3 log x.x4 dx= elog x3·x4 dx         alogx=logxa= x3·x4 dx                elog m=m=x7·dx=x88+C

Page No 19.197:

Question 3:

Write a value of x2 sin x3 dx.

Answer:

Let I=x2.sin x3 dx
Let x3t
⇒ â€‹3x2dx = dt
x2 dx=dt3I=13 sin t dt     =13-cos t+C     =-13cos x3+C       t=x3

Page No 19.197:

Question 4:

Write a value of tan3 x sec2 x dx.

Answer:

Let I= tan3 x . sec2 x dx
Let tan x = t
⇒ sec2x dx = dt
I= t3 . dt
    =t44+C=tan4 x4+C             t=tan x

Page No 19.197:

Question 5:

Write a value of ex sin x+cos x dx.

Answer:

Let I= ex (sin x + cos x) dx
Let ex sin x = t
⇒ (ex . sin x + ex cos x) dx = dt
 I=dt       =t+C       = exsin x+C      t=ex sinx

Page No 19.197:

Question 6:

Write a value of tan6 x sec2 x dx.

Answer:

Let I= tan6 x . sec2 x dx
Let tan x = t
sec2 x dx = dt
I= t6 . dt
    =t77+C=tan7 x7+C    t=tan x

Page No 19.197:

Question 7:

Write a value of cos x3+2 sin x dx.

Answer:

Let I= cos x3+2 sin xdxLet 3+2 sin x=t2 cos x dx=dtcos x dx=dt2I=12dtt     =12 log t+C    =12log 3+2 sin x+C      t=3+2 sinx 

Page No 19.197:

Question 8:

Write a value of ex sec x 1+tan x dx.

Answer:

Let I= ex sec x(1 + tan x) dx
       = ex (sec x + sec x tan x) dx
Let ex sec x = t
⇒ (ex sec x + ex sec x tan x)dx = dt
⇒​ ex sec x (1 + tan x) dx = dt
I= dt
     = t + C
     = ex sec x + C                 t=exsec x

Page No 19.197:

Question 9:

Write a value of log xnx dx.

Answer:

Let I= log xnx dx       = n log xxdx       log xa=a logxLet log x=t1xdx=dtI=n  t dt     =n.t22+C     =n.log x22+C         t=log x

Page No 19.197:

Question 10:

Write a value of log xnx dx.

Answer:

Let I=log xnxdxLet log x=t1xdx=dtI= tn dt     =tn+1n+1+C      =log xn+1n+1+C        t=log x

Page No 19.197:

Question 11:

Write a value of elog sin x cos x dx.

Answer:

Let I= elog sin x . cos x dx
       = sin x × cos x dx             elog a=a
Let sin x = t
⇒​ cos x dx = dt
I= t . dt
    =t22+C=sin2 x2+C       t= sin x 

Page No 19.197:

Question 12:

Write a value of sin3 x cos x dx.

Answer:

Let I= sin3 x . cos x dx
Let sin x = t
⇒​ cos x dx = dt
 I=​  t3 . dt
    =t44+C=sin4 x4+C       t=sin x

Page No 19.197:

Question 13:

Write a value of cos4 x sin x dx.

Answer:

Let I=cos4 x .sin x dx
Let cos x = t
⇒​ –sin x dx = dt
⇒ sin x dx = –dt
 I= –​​ t4 dt
     =-t55+C=-cos5 x5+C     t= cos x

Page No 19.197:

Question 14:

Write a value of tan x sec3 x dx.

Answer:

Let I=tan x . sec3x dx
       =sec2 x . sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
I=t2 dt
    =t33+C=sec3 x3+C      x=sec x

Page No 19.197:

Question 15:

Write a value of 11+ex dx.

Answer:

Let I=dx1+exDividing numerator & denominator by exI=1exdx1ex+1      =e-x dxe-x +1Let e-x+1=t-e-x dx=dte-x dx=-dtI=-dtt      =-log t+C      =-log 1+ex+C      t=1+ex

Page No 19.197:

Question 16:

Write a value of 11+2ex dx.

Answer:

Let I=dx1+2exDividing numerator & denominator by exI=1exdx1ex+2     =e-x dxe-x+2Let e-x+2=t-e-x dx=dte-x dx=-dtI=-dtt     =-log t+C     =-log e-x+2+C        t=e-x+2

Page No 19.197:

Question 17:

Write a value of tan-1 x31+x2 dx.

Answer:

Let I=tan-1 x31+x2dxLet tan-1 x=t11+x2dx=dtI= t3.dt     =t44+C     =tan-1 x44+C      t=tan-1 x

Page No 19.197:

Question 18:

Write a value of sec2 x5+tan x4 dx.

Answer:

Let I=sec2 x dx5+tan x4Let 5+tan x=t sec2 x dx=dtI=dtt4    = t-4 dt   =t-4+1-4+1+C   =-13t3+C   =-13 5t+tan x3+C        t=5+tan x

Page No 19.197:

Question 19:

Write a value of sin x+cos x1+sin 2x dx.

Answer:

 sin x+cos x1+sin 2xdx sin x+cos xdxsin2 x+cos2 x+2 sin x cos x sin x+cos xdxsin x+cos x2 dxx+C

Page No 19.197:

Question 20:

Write a value of loge x dx.

Answer:

loge x dx
=  1II.loge xI dx
=loge x1 dx-ddxloge x1 dxdx
= loge x 1 . dx – â€‹ 1x×x.dx
= loge x × x – â€‹ dx
=​ x loge xx + C
=​ x loge xx + C
= x (loge x – 1) + C

Page No 19.197:

Question 21:

Write a value of ax ex dx.

Answer:

ax . ex dx
= â€‹∫ (ae)x dx
=aexln ae+C

Page No 19.197:

Question 22:

Write a value of e2x2+ln x dx.

Answer:

Let I= e2x2+ln xdx= e2x2×eln xdx= e2x2. x dxLet 2x2=t4x dx=dtx dx=dt4I=14 et dt     =14 et+C     =14e2x2+C        t=2x2

Page No 19.197:

Question 23:

Write a value of ex loge a+ea loge x dx.

Answer:

ex loge a+ea loge xdx elog ax+elog xa dx= ax+xadx=axlog a+xa+1a+1+C

Page No 19.197:

Question 24:

Write a value of cos xsin x log sin x dx.

Answer:

Let I= cos xsin x ·log sin xdx cot xlog sin xdxLet log sin x=tcot x dx=dtI= dtt     =log t+C     =log log sin x+C

Page No 19.197:

Question 25:

Write a value of sin 2xa2 sin2 x+b2 cos2 x dx.

Answer:

Let I= sin 2x dxa2 sin2+b2 cos2 xLet a2 sin2 x+b2 cos2 x=ta22 sin x cos x+b22 cos x×-sin xdx=dta2-b2 sin 2x.dx=dtsin 2x dx=dta2-b2I=1a2-b2dtt     =1a2-b2log t+C    =1a2-b2log a2 sin2x+b2 cos2 x+C         t=a2 sin2 x+b2 cos2 x

Page No 19.197:

Question 26:

Write a value of ax3+ax dx.

Answer:

Let I= ax dx3+axLet 3+ax=tax.log a dx=dtax dx=dtlog aI=1log adtt     =1log alog t+C     =1log alog 3+ax+C                ∵ t=3+ax    

Page No 19.197:

Question 27:

Write a value of 1+log x3+x log x dx.

Answer:

Let I= 1+log x3+x log xdxLet 3+x log x=t0+x.1x+log xdx=dt1+log xdx=dtI= dtt     =log t+C     =log 3+x log x+C        t=3+x log x

Page No 19.197:

Question 28:

Write a value of sin xcos3 x dx.

Answer:

Let I= sin xcos3 xdxLet cos x=t-sin x dx=dtsin x dx=-dtI=- dtt3     =- t-3 dt     =-t-3+1-3+1+C     =12t2+C     =12 cos2 x+C       t=cos x     =12 sec2 x+C    

Page No 19.197:

Question 29:

Write a value of sin x-cos x1+sin 2x dx.

Answer:

Let I = sin x+cos x dx1-sin 2x         =sin x+cos x dxsin2x+cos2x-2 sin x cos x         =sin x+cos x dxsin x-cos x2         =sin x+cos x dxsin x-cos x         =±sin x+cos xsin x-cos xdxLet sin x-cos x=tcos x+sin xdx=dt I=±dtt       =±ln t+C       =±ln sin x-cos x+C        t=sin x-cos x

Page No 19.197:

Question 30:

Write a value of 1x log xn dx.

Answer:

Let I= dxxlog xn       = log x-n dxxLet log x=t1xdx=dtI= t-n.dt      =t-n+1-n+1+C      =log x-n+1-n+1+C             t=log x

Page No 19.197:

Question 31:

Write a value of eax sin bx dx.

Answer:

Let I= eax.sin bx dx        =sin bxeaxdx-ddxsin bxeaxdxdx       =sin bx×eaxa-cos bx×b.eaxa       =sin bx×eaxa-baeax.cos bx dx        =sin bx×eaxa-baI1      ...1I1= eax ×cos bxdx       =cos bxeaxdx-ddxcos bxeaxdxdx      =cos bx×eaxa+b.sin bx×eaxadx      =cos bx.eaxa+baI        ....2From 1 & 2I=sin bx.eaxa-ba cos bx.eaxa+baII =sin bx.eaxa-ba2 cos bx eax-b2a2II+b2a2I=sin bx.eaxa-b cos bx eaxa2a2+b2I=a sin bx-bcos bxeaxI=a sin bx-bcos bx eaxa2+b2+C

Page No 19.197:

Question 32:

Write a value of eax cos bx dx.

Answer:

Let I= eax.cos bx dx        =cos bxeaxdx-ddxcos bxeaxdxdx       =cos bx×eaxa--sin bx×b.eaxa       =cos bx×eaxa+baeax.sin bx dx        =cos bx×eaxa+baI1      ...1I1= eax ×sin bxdx       =sin bxeaxdx-ddxsin bxeaxdxdx      =sin bx×eaxa-b.cos bx×eaxadx      =sin bx.eaxa-baI        ....2From 1 & 2I=cos bx.eaxa+ba sin bx.eaxa-baII =cos bx.eaxa+ba2 sin bx eax-b2a2II+b2a2I=cos bx.eaxa+b sin bx eaxa2a2+b2I=a cos bx+b sin bxeaxI=a cos bx+bsin bx eaxa2+b2+C



Page No 19.198:

Question 33:

Write a value of ex 1x-1x2 dx.

Answer:

Let I = ex 1x-1x2dxAs we know that  ex fx+f'xdx=ex fx+CI=exx+C

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Question 34:

Write a value of eax a fx+f'x dx.

Answer:

Let I = eax a fx+f'xdxLet eax.fx=teax.a fx+eax.f'xdx=dtI= dt     =t+C     =eax.fx+C      t=eax.fx

Page No 19.198:

Question 35:

Write a value of 4-x2 dx.

Answer:

 4-x2 dx= 22-x2dx=x222-x2+222sin-1x2+C      a2-x2=x2a2-x2-a22sin-1xa+C=x24-x2+2 sin-1 x2+C

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Question 36:

Write a value of 9+x2 dx.

Answer:

 9+x2 dx= 32+x2 dx           a2+x2=x2x2+a2+a22ln x+x2+a2=x29+x2+92ln x+9+x2+C

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Question 37:

Write a value of x2-9 dx.

Answer:

 x2-9 dx= x2-32 dx=x2x2-32-322ln x+x2-3+C        x2-a2=x2x2-a2-a22ln x+x2+a2+C=x2x2-9-92ln x+x2-9+C

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Question 38:

Evaluate: x21+x3 dx

Answer:

Let I= x2 dx1+x3Putting 1+x3=t3x2 dx=dtx2 dx=dt3I=13 dtt     =13ln t+C     =13ln 1+x3+C          t=1+x3

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Question 39:

Evaluate: x2+4xx3+6x2+5 dx

Answer:

Let I=x2+4xx3+6x2+5 dxLet x3+6x2+5=t3x2+12x dx=dtx2+4x dx=dt3Putting x3+6x2+5=t and x2+4x dx=dt3I=13dtt      =13 ln  t+C      =13ln x3+6x2+5+C

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Question 40:

Evaluate: sec2xx dx

Answer:

Let I =sec2xx dxLet x=tdx2x=dtdxx=2 dtPutting  x=t and dxx=2 dtI=2sec2+dt     =2 tan t+C     =2 tan x+C        t=x

Page No 19.198:

Question 41:

Evaluate: sin xx dx

Answer:

Let I=sinxx dxLet  x=t12xdx=dtdxx=2 dtPutting x=t and dxx=2 dt , we get I=2sin t dt     =-2 cos t+C            t=x     =-2 cos x+C    

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Question 42:

Evaluate: cos xx dx

Answer:

Let I=cosxx dxPutting  x=t12xdx=dtdxx=2 dt I=2cos t dt      =2 sin t+C    , where t=x      =2 sin x+C     

Page No 19.198:

Question 43:

Evaluate: 1+log x2x dx

Answer:

Let 1+log x2x dxPutting 1+log x=t1x dx=dtI=t2·dt     =t33+C     =1+log x33+C          t=1+log  x

Page No 19.198:

Question 44:

Evaluate: sec2 7-4x dx

Answer:

sec2 7-4x dx=tan 7-4x-4+C        sec2 x=tan x+C

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Question 45:

Evaluate: log xx dx

Answer:

Let I=log xx dx& let log x=t1x dx=dtI=t·dt     =t22+C     =log x22+C       t=log x

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Question 46:

Evaluate: 2x dx

Answer:

2x dx=2xln 2+C           ax dx=axln a+C

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Question 47:

Write a value of 1-sin xcos2 x dx

Answer:

    1-sin xcos2x dx=1cos2x-sin xcos2xdx    1cos2x-sin xcos x×1cos x dx=sec2x-sec x tan x dx=tan x-sec x+C

Page No 19.198:

Question 48:

Evaluate: x3-1x2 dx

Answer:

    x3-1x2 dx=x3x2-1x2dx=x-x-2dx=x22-x-2+1-2+1+C=x22+1x+C

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Question 49:

Evaluate: x3-x2+x-1x-1 dx

Answer:

    x3-x2+x-1x-1 dx=x2 x-1+1x-1x-1dx=x2+1 x-1x-1dx=x2+1 dx=x33+x+C

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Question 50:

Evaluate: etan-1 x1+x2 dx

Answer:

Let I=etan-1x1+x2 dxLet tan-1x=tdx1+x2=dtI=etdt     =et+C     =etan-1x+C

Page No 19.198:

Question 51:

Evaluate: 11-x2 dx

Answer:

Let I=dx1-x2Let x=sin θdx=cos θ I=cos θ cos θdθ        =dθ        =θ+C        =sin-1x+C      x=sin θ

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Question 52:

Write the value of sec x sec x+tan x dx

Answer:

sec x sec x+tan x dx=sec2x+sec x tan x dx=tan x+sec x+C                          

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Question 53:

Evaluate: 1x2+16dx

Answer:

1x2+16dx=1x2+42dx                   =14tan-1x4+cHence, 1x2+16dx=14tan-1x4+c

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Question 54:

Evaluate: 1-xx dx

Answer:

1-xx dx=x-xx dx                       =x12-x32 dx                       =x12+112+1-x32+132+1+c                       =23x32-25x52+cHence, 1-xx dx=23x32-25x52+c

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Question 55:

Evaluate: x+cos6x3x2+sin6xdx

Answer:

x+cos6x3x2+sin6xdx   Let 3x2+sin6x=t   On differentiating both sides, we get   6x+6cos6x dx=dtx+cos6x3x2+sin6xdx=161tdt                             =16logt+c                             =16log3x2+sin6x+cHence, x+cos6x3x2+sin6xdx=16log3x2+sin6x+c

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Question 56:

If x-1x2ex dx=fxex+C, then write the value of fx.

Answer:

x-1x2ex dx=xx2-1x2ex dx                       =1x-1x2ex dxConsider, fx=1x, then f'x=-1x2Thus, the given integrand is of the form exfx+f'x.Therefore, x-1x2ex dx=1xex+CHence, fx=1x.

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Question 57:

If extanx+1secx dx=ex fx+C, then write the value of fx.

Answer:

ex tanx+1 secx dx=ex tanxsecx+secx dx                                 =ex secx+tanxsecx dxConsider, fx=secx, then f'x=tanxsecxThus, the given integrand is of the form exfx+f'x.Therefore, ex tanx+1 secx dx=secx ex+CHence, fx=secx.

Page No 19.198:

Question 58:

Evaluate: 21-cos2xdx

Answer:

21-cos2xdx=22sin2xdx                       =cosec2x dx                       =-cotx+cHence, 21-cos2xdx=-cotx+c.

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Question 59:

Write the anti-derivative of 3x+1x.
 

Answer:

3x+1xdx=3 x12+112+1+x-12+1-12+1+c                            =2x32+2x12+c                            =2x32+x12+cHence, the anti-derivative of 3x+1x is 2x32+x12+c.

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Question 60:

Evaluate: cos-1sinx dx

Answer:

cos-1sinx dx=cos-1cosπ2-x dx                        =π2-x dx                        =π2x-12x2+cHence, cos-1sinx dx=π2x-12x2+c

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Question 61:

Evaluate: 1sin2x cos2xdx

Answer:

1sin2x cos2xdx=414 sin2x cos2xdx                         =41sin22xdx                         =4cosec22x dx                         =-42cot2x+c                         =-2cot2x+cHence, 1sin2x cos2xdx=-2cot2x+c

Page No 19.198:

Question 62:

Evaluate : 1x(1+logx) dx

Answer:

I1x1+logx dx

Let (1 + log x) = t

or, 1xdx=dtI=1tdt

I=logt+C I=log1+log x+C



Page No 19.199:

Question 1:

x4+x4 dx is equal to

(a) 14tan-1 x2+C

(b) 14tan-1 x22

(c) 12tan-1 x22

(d) none of these

Answer:

(b) 14tan-1 x22

Let I=x4+x4dx        =x dx22+x22Putting x2=t2x dx=dtx dx=dt2I=12dt22+t2     =12×12 tan-1 t2+C                      1a2+x2=1atan-1xa     =14 tan-1 x22+C                             t =x2

Page No 19.199:

Question 2:

1cos x+3 sin x dx is equal to

(a) log tanπ3+x2+C

(b) log tan x2-π3+C

(c) 12 log tan x2+π3+C

(d) none of these

Answer:

(d) none of these

1cos x+ 3 sin xdx=12dxcos x×12+sin x×32=12dxcos x·cosπ3+sin x·sinπ3=12dxcos x-π3=12sec x-π3dx=12ln tan π4+12x-π3+C=12ln tan π4+x2-π6+C=12ln tan x2+π12+C
 



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Question 3:

x sec x2 dx is equal to

(a) 12 log (sec x2 + tan x2) + C

(b) x22 log (sec x2 + tan x2) + C

(c) 2 log (sec x2 + tan x2) + C

(d) none of these

Answer:

(a) 12 log (sec x2 + tan x2) + C

Let I=x sec x2 dxPutting x2=t2x dx=dtx dx=dt2 I=12sec t·dt       =12 log sec t+tan t+C       =12 log sec x2+tan x2+C           t= x2

 

Page No 19.200:

Question 4:

If 15+4 sin x dx=A tan-1 B tanx2+43+C, then

(a) A = 23, B = 53

(b) A = 13, B = 23

(c) A = -23, B = 53

(d) A = 13, B = -53

Answer:

(a) A = 23, B = 53


15+4 sin xdx=A tan-1 B tan x2+43+C           ....(1)Considering the LHS of eq (1)Putting  sin x=2 tan x21+tan2 x215+8 tan x21+tan2 x2dx1+tan2 x25 1+tan2 x2+8 tan x2dxsec2 x25 tan2 x2+8 tan x2+5 dx       ...(2) Let tan x2=tsec2 x2×12 dx=dtsec2 x2 dx=2dt Putting tan x2=t and sec2 x2 dx=2dt we get, 2dt5t2+8t+525dtt2+85t+125dtt2+85t+452-452+125dtt+452+1-162525dtt+452+35225×53 tan-1 t+4535+C23 tan-1 5t+43+C23 tan-1 53 tan x2+43+C                t= tan x2      ...(3)Comparing eq (3) with the RHS of eq (1) we get ,  A=23, B=53
 

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Question 5:

xsin x sin xx+cos x . log x dx is equal to

(a) xsin x + C

(b) xsin x cos x + C

(c) xsin x22+C

(d) none of these

Answer:

(a) xsin x + C

Let I=xsin x sin xx+cos x·log xdxPutting  xsin x=tln xsin x=ln tsin x·ln x=ln tsin x×1x+cos x ln xdx=1tdtI=t·dtt     =t+C     =xsin x+C        t= xsin x
 

Page No 19.200:

Question 6:

Integration of 11+loge x2 with respect to loge x is
(a) tan-1 loge xx+C

(b) tan-1 loge x+C

(c) tan-1 xx+C

(d) none of these

Answer:

(b) tan-1 loge x+C

We have to integrate 11+logex2 with respect to log ex  Let I=d loge x1+loge x2Putting  loge x=td loge x=dt I=dt1+t2      =tan-1 t+C      =tan-1 loge x+C         t=logex

 

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Question 7:

If cos 8x+1tan 2x-cot 2x dx = a cos 8x + C, then a =

(a) -116

(b) 18

(c) 116

(d) -18

Answer:

(c) 116

If cos 8x+1tan 2x-cot 2xdx =a cos 8x+C        ...(1) Considering the LHS of eq (1)cos 8x+1tan 2x-cot 2xdx2 cos24xsin 2xcos 2x-cos 2xsin 2xdx2 cos2 4xsin2 2x-cos2 2x×sin 2x cos 2x-cos2 4x×2 sin 2x·cos 2xcos2 2x-sin2 2xdx-cos2 4x×sin 4xcos 4xdx                    cos 2x= cos2x-sin2x12-2 sin 4x cos 4x dx         -1  2sin 8x dx-12- cos 8x8+C=116cos 8x+C             ....(2) Comparing RHS of eq (1) with the eq (2) a=116
 

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Question 8:

If sin8 x-cos8 x1-2 sin2 x cos2 x dx = a sin 2x + C, then a =

(a) −1/2
(b) 1/2
(c) −1
(d) 1

Answer:

(a) −1/2


 If sin8x-cos8x1-2 sin2x cos2xdx=a sin 2x+C     ....(1)Considering LHS of eq (1) sin4x-cos4x sin4x+cos4x1-2 sin2xcos2xsin2x-cos2x sin2x+cos2x·sin4x+cos4x dxsin2x+cos2x2-2 sin2x cos2xsin2x-cos2x·sin4x+cos4xdxsin4x+cos4x+2 sin2x cos2x-2 sin2x cos2x-cos2x-sin2x×sin4x+cos4x dxsin4x+cos4x-cos 2x dx                    cos2x-sin2x=cos 2x           ...(2)Comparing the RHS of eq (1) with eq (2) we get , a=-12
 

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Question 9:

x-1 e-x dx is equal to

(a) − xex + C

(b) xex + C

(c) − xex + C

(d) xex + C

Answer:

x-1I e-xII dx=x-1e-x dx-ddxx-1e-x dxdx=x-1·e-x -1-1·e-x×-1 dx=-x-1 e-x+e-x dx=1-x e-x+e-x-1+C1-x-1 e-x+C=-x e-x+C
 

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Question 10:

If 21/xx2 dx=k21/x+C, then k is equal to

(a) -1loge 2

(b) − loge 2

(c) − 1

(d) 12

Answer:

(a) -1loge 2

If21xx2dx=k·21x+C       ....(1) Let  1x=t-1x2dx=dtdxx2=-dtPutting 1x=t and dxx2=-dt in LHS of eq (1) , we get -2t·dt-2tln 2+C-21xln 2+C       ...(2) Comparing RHS of eq (1) with eq (2) we get ,  k=-1ln 2 or  -1loge2
 

Page No 19.200:

Question 11:

11+tan x dx=

(a) loge (x + sin x) + C

(b) loge (sin x + cos x) + C

(c) 2 sec2x2+C

(d) 12 [x + log (sin x + cos x)] + C

Answer:

Disclaimer : Generally here book is taking loge x  as log x . So we are writing ln x or loge x instead log x only .
(d) 12 [x + ln (sin x + cos x)] + C

Let I=11+tan xdx       =11+sin xcos xdx       =cos x cos x+sin xdx       =122 cos x cos x +sin xdx       =12cos x+sin x+cos x-sin xcos x+sin xdx       =12cos x +sin xcos x+sin xdx+12cos x-sin xcos x+sin xdx       =12dx+12cos x-sin xcos x+ sin xdxPutting  sin x+cos x=tcos x-sin x dx=dt I=12dx+12dtt       =x2+12ln t+C       =x2+12 ln cos x+sin x+C        t=sin x+cos x      =12x+ln sin x+ cos x+C  
 

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Question 12:

x3 dx is equal to

(a) -x44+C

(b) x44+C

(c) x44+C

(d) none of these

Answer:

(d) none of these

x3 dxx=x, x0-x, x<0Case 1:When x0 x3 dx=x3 dx=x44+CCase 2:x<0x3 dx=-x3 dx=-x44+C
 

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Question 13:

The value of cos xx dx is
(a) 2 cos x + C

(b) cos xx+C

(c) sin x+C

(d) 2 sin x+C

Answer:

(d) 2 sin x+C

Let I=cos xxdxPutting  x=t12xdx=dtdxx=2dt I=2cos t·dt      =2 sin t+C      =2 sin x+C                  t=x



Page No 19.201:

Question 14:

ex 1-cot x+cot2 x dx=
(a) ex cot x + C
(b) −ex cot x + C
(c) ex cosec x + C
(d) −ex cosec x + C

Answer:

(b) −ex cot x + C

Let I=ex1-cot x+cot2xdx       =excosec2x-cot xdxAs we know that efx+f'xx=exfx+C I=-ex cot x+C

 

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Question 15:

sin6 xcos8 x dx=

(a) tan 7x + C

(b) tan7 x7+C

(c) tan 7x7+C

(d) sec7 x + C

Answer:

(b) tan7 x7+C

Let I=sin6x cos8xdx       =sin6xcos6x×1cos2xdx      =tan6x·sec2x dxPutting  tan x=tsec2x dx=dtI=t6·dt      =t77+C      =tan7x7+C         t= tan x

 

Page No 19.201:

Question 16:

17+5 cos x dx=

(a) 16tan-116tanx2+C

(b) 13tan-113tanx2+C

(c) 14tan-1tanx2+C

(d) 17tan-1tanx2+C

Answer:

(a) 16tan-116tanx2+C

Let I=dx7+5 cos xPutting cos x =1-tan2 x21+tan2 x2I=dx7+5×1-tan2 x21+tan2 x2     =1+tan2 x2 dx71+tan2 x2+5-5 tan2 x2    =sec2 x2 dx2 tan2 x2+12    =12sec2x2dxtan2x2+62Let tan x2=t12 sec2 x2 dx=dtsec2 x2 dx=2 dtI=122 dtt2+62     =16 tan-1 t6+C            1a2+x2=1atan-1xa+C     =16 tan-1 tan x26+C         t= tan x2
 

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Question 17:

11-cos x-sin x dx=

(a) log1+cotx2+C

(b) log1-tanx2+C

(c) log1-cotx2+C

(d) log1+tanx2+C

Answer:

(c) log1-cotx2+C
Disclaimer : Here in answer log1-cotx2+C refers to loge1- cotx2+C or ln 1-cotx2+C

Let I=dx1-cos x-sin x        =dx1-1-tan2 x21+tan2 x2-2 tan x21+tan2 x2        =1+tan2 x2 dx1+tan2 x2-1-tan2 x2-2 tan x2        =sec2 x2 dx2 tan2 x2-2 tan x2        =12sec2 x2 dxtan2 x2-tan x2Putting tan x2=t12sec2 x2 dx=dtsec2 x2 dx=2dtI=122dtt2-t      =dtt2-t+122-122      =dtt-122-122      =12×12 ln t-12-12t-12+12+C       dxx2-a2=12alnx-ax+a+C      =ln t-1t+C      =ln 1-1t+C      =ln 1-1tan x2+C                        t= tan x2      =ln 1-cot x2+C 

 

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Question 18:

x+3x+42ex dx=

(a) exx+4+C

(b) exx+3+C

(c) 1x+42+C

(d) exx+42+C

Answer:

(a) exx+4+C

Let I=x+3x+42ex dxx+4-1x+42 ex dx1x+4-1x+42 ex dxAs, we know that exfx+f'x dx=exfx+C I=exx+4+C
 

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Question 19:

sin x3+4 cos2 x dx

(a) log (3 + 4 cos2 x) + C

(b) 12 3tan-1cos x3+C

(c) -12 3tan-12 cos x3+C

(d) 12 3tan-12 cos x3+C

Answer:

(c) -12 3tan-12 cos x3+C

Let I=sin x 3+4 cos2xdxPutting  cos x=t-sin x dx=dtI=-dt3+4t2     =14-dtt2+322     =-14×132 tan-1 t×23+C                  1x2+a2=1atan-1xa+C     =-123 tan-1 2 t3+C     =-123 tan-1 2 cos x3+C          t=cos x

Page No 19.201:

Question 20:

ex 1-sin x1-cos x dx

(a) -ex tanx2+C




(c) -12ex tanx2+C

(d) -12ex cotx2+C

Answer:

(b) -ex cotx2+C

Let I =ex1-sin x1-cos xdxex11-cos x-sin x1-cos xdxex 12 sin2 x2-2 sin x2 cos x22 sin2 x2dxex 12 cosec2 x2-cot x2dxAs, we know that exfx+f'x dx=exfx+C I=-ex cot x2+C

 

Page No 19.201:

Question 21:

2ex+e-x2 dx

(a) -e-xex+e-x+C

(b) -1ex+e-x+C

(c) -1ex+12+C

(d) 1ex-e-x+C

Answer:

(a) -e-xex+e-x+C

Let I =2 dxex+e-x2        =2 dxex+1ex2        =2e2x dxe2x+12Let e2x+1=te2x·2 dx=dte2x·dx=dt2I=2×12dtt2     =-1t+C     =-1e2x+1+C         t=e2x+1

Dividing numerator and denominator by ex
I=-1exex+1ex      =-e-xex+e-x+C
 

Page No 19.201:

Question 22:

ex 1+xcos2 xex dx=

(a) 2 loge cos (xex) + C

(b) sec (xex) + C

(c) tan (xex) + C

(d) tan (x + ex) + C

Answer:

(c) tan (xex) + C

Let I=ex1+xcos2 xexdxPutting  xex=t1·ex+x exdx=dtex1+xdx=dtI=dtcos2 t     =sec2t dt     =tan t+C     =tan x ex+C        t= xex
 

Page No 19.201:

Question 23:

sin2 xcos4 x dx=

(a) 13tan2 x+C

(b) 12tan2 x+C

(c) 13tan3 x+C

(d) none of these

Answer:

(c) 13tan3 x+C

Let I =sin2x dxcos4x        =sin2xcos2x×1cos2xdx        =tan2x·sec2x dxLet tan x=tsec2x dx=dt I=t2·dt     =t33+C     =tan3x3+C               t=tan x



Page No 19.202:

Question 24:

The primitive of the function
fx=1-1x2 ax+1x, a>0 is

(a) ax+1xloge a

(b) loge a·ax+1x

(c) ax+1xxloge a

(d) xax+1xloge a

Answer:

(a) ax+1xloge a

fx=1-1x2·ax+1xfxdx=1-1x2·ax+1xdxLet x+1x=t1-1x2dx=dtfxdx=at·dt                  =atloge a+C                  =ax+1xloge a+C          t =x+1x
 

Page No 19.202:

Question 25:

The value of 1x+x log x dx is
(a) 1 + log x
(b) x + log x
(c) x log (1 + log x)
(d) log (1 + log x)

Answer:

(d) log (1 + log x)

Let I=dxx+x log xdxx 1+log xPutting 1+log x=t1x dx=dt I=dtt      =ln t+C      =ln 1+log x+C
 

Page No 19.202:

Question 26:

x1-x dx is equal to
(a) sin-1 x+C

(b) sin-1 x-x 1-x+C

(c) sin-1 x 1-x+C

(d) sin-1 x-x 1-x+C

Answer:

(d) sin-1 x-x 1-x+C

Let I =x1-xdxPutting  x=sin θx=sin2θdx=2 sin θ cos θ dθdx=sin 2θ dθ I=sin2θ1-sin2θ×sin 2θ·dθ      =sin θcos θ×2 sin θ·cos θ dθ      =2 sin2θ·dθ      =1-cos 2θdθ      =θ-sin 2θ2+C      =θ-2 sin θ cos θ2+C       =θ-sin θ 1-sin2θ+C      =sin-1 x-x 1-x+C           θ=sin-1x     =sin-1 x-x1-x+C   

 

Page No 19.202:

Question 27:

ex fx+f'x dx=

(a) ex f (x) + C
(b) ex + f (x)
(c) 2ex f (x)
(d) exf (x)

Answer:

(a) ex f (x) + C

Let I =exfx+f'xdxPutting exfx=tex·fx+exf'xdx=dtI=dt      =t+C      =exfx+C      t=exfx

Page No 19.202:

Question 28:

The value of sin x+cos x1-sin 2x dx is equal to
(a) sin 2x+C
(b) cos 2x+C
(c) ± (sin x − cos x) + C
(d) ± log (sin x − cos x) + C

Answer:

(d) ± log (sin x − cos x) + C

 Let I = sin x+cos x dx1-sin 2x         =sin x+cos x dxsin2x+cos2x-2 sin x cos x         =sin x+cos x dxsin x-cos x2         =sin x+cos x dxsin x-cos x         =±sin x+cos xsin x-cos xdxLet sin x-cos x=tcos x+sin xdx=dt I=±dtt       =±ln t+C       =±ln sin x-cos x+C        t=sin x-cos x

Page No 19.202:

Question 29:

If x sin x dx = −x cos x + α, then α is equal to
(a) sin x + C
(b) cos x + C
(c) C
(d) none of these

Answer:

(a) sin x + C

xI·sin xII dx=-x cos x+α xsin x dx-ddxxsin x dxdx=-x cos x+αx -cos x-1·-cos xdx=-x cos x+α-x cos x+cos x dx=-x cos x+α-x cos x +sin x+C=-x cos x+α α=sin x+C
 

Page No 19.202:

Question 30:

cos 2x-1cos 2x+1 dx=
(a) tan xx + C
(b) x + tan x + C
(c) x − tan x + C
(d) − x − cot x + C

Answer:

(c) x − tan x + C

cos 2x-1cos 2x+1dx=1-2 sin2x-12 cos2x-1+1dx=-tan2x dx=-sec2x-1dx=1-sec2xdx=x-tan x+C
 

Page No 19.202:

Question 31:

cos2x-cos2θcosx-cosθdx is equal to

(a) 2sinx+xcosθ+C
(b) 2sinx-xcosθ+C
(c) 2sinx+2xcosθ+C
(d) 2sinx-2xcosθ+C

Answer:

Let I=cos2x-cos2θcosx-cosθdx       =2cos2x-1-2cos2θ-1cosx-cosθdx       =2cos2x-1-2cos2θ+1cosx-cosθdx       =2cosx-cosθcosx+cosθcosx-cosθdx       =2cosx+cosθdx       =2sinx+xcosθ+CTherefore, cos2x-cos2θcosx-cosθdx=2sinx+xcosθ+CHence, the correct option is (a).

Page No 19.202:

Question 32:

x94x2+16dx is equal to

(a) 15x4+1x2-5+C(b) 154+1x2-5+C(c) 110x1x2+4-5+C(d) 1101x2+4-5+C

Answer:

Let I=x94x2+16dx       =x9x124+1x26dx       =1x34+1x26dx        Let 4+1x2=t        On differentiating both sides, we get        -2x3dx=dt I=-121t6dt      =-12-15t-5+C      =1104+1x2-5+CTherefore, x94x2+16dx=1104+1x2-5+CHence, the correct option is (d).

Page No 19.202:

Question 33:

x31+x2dx=a1+x232+b1+x2+C, then

(a) a=13, b=1(b) a=-13, b=1(c) a=-13, b=-1(d) a=13, b=-1

Answer:

Let I=x31+x2dx       =x.x21+x2dx               Let 1+x2=t        On differentiating both sides, we get        2x dx=dt I=12t-1tdt      =12tt-1tdt      =12t12-t-12dt      =1223t32-21t12+C      =131+x232-1+x2+CSince, x31+x2dx=a1+x232+b1+x2+CTherefore, a=13, b=-1.Hence, the correct option is (d).

Page No 19.202:

Question 34:

x3x+1dx is equal to

(a) x+x22+x33-log1-x+C(b) x+x22-x33-log1-x+C(c) x-x22-x33-log1+x+C(d) x-x22+x33-log1+x+C

Answer:

Let I=x3x+1dx       =x3+1-1x+1dx       =x3+1x+1-1x+1dx       =x+1x2-x+1x+1-1x+1dx       =x2-x+1-1x+1dx       =x33-x22+x-logx+1+C       =x33-x22+x-logx+1+CTherefore, x3x+1dx=x33-x22+x-logx+1+CHence, the correct option is (d).



Page No 19.203:

Question 35:

If 1x+2x2+1dx=alog1+x2+btan-1x+15logx+2+C, then

(a) a=-110, b=-25(b) a=110, b=-25(c) a=-110, b=25(d) a=110, b=25

Answer:

Let I=1x+2x2+1dxWe express,1x+2x2+1=Ax+2+Bx+Cx2+11=Ax2+1+Bx+Cx+2On comparing the coefficients of x2, x and constants, we get0=A+B     and     0=2B+C     and     1=A+2Cor A=15     and     B=-15     and     C=25 I=15x+2+-15x+25x2+1dx      =151x+2dx-15xx2+1dx+251x2+1dx      =15logx+2-110logx2+1+25tan-1x+CSince, 1x+2x2+1dx=alog1+x2+btan-1x+15logx+2+CTherefore, a=-110 and b=25.Hence, the correct option is (c).

Page No 19.203:

Question 1:

1x+x+1 dx

Answer:

1x+1+xdxRationalising the denominator, =x+1-xx+1+x x+1-xdx=x+1-xx+1-xdx=x+1-x dx=x+112-x12dx=x+112+112+1-x12+112+1+C=23 x+132-23 x32+C=23x+132- x32+C

Page No 19.203:

Question 2:

1-x41-x dx

Answer:

1-x41-xdx=1-x2 1+x21-xdx=1-x 1+x 1+x21-xdx=1+x 1+x2dx=1+x2+x+x3dx=x+x33+x22+x44+C

Page No 19.203:

Question 3:

x+2x+13 dx

Answer:

Let x+2x+13dxPutting  x+1=tx=t-1dx=dtI=t-1+2t3dt     =1t2+1t3dt     =t-2+t-3dt     =t-2+1-2+1+t-3+1-3+1+C       =-1t-2t2+C     =-1x+1-12x+12+C         t=x+1

Page No 19.203:

Question 4:

8x+134x+7 dx

Answer:

Let I=8x+134x+7dxPutting  4x+7=tx=t-744dx=dtdx=dt4 I=148 t-74+13tdt      =142t-14+13tdt      =142t-1tdt      =142ttdt-14dtt      =12t12dt-14t-12dt      =12t12+112+1-14t-12+1-12+1+C      =12×23t32-24 t12+C      =13 t32-24t12+C     =13 4x+732-12 4x+712+C               t=4x+7      =13 4x+732-4x+7+C

Page No 19.203:

Question 5:

1+x+x2x2 1+x dx

Answer:

 1+x+x2x2 1+xdx1+xx2 1+xdx+x2x2 1+xdxdxx2+dx1+xx-2dx+11+xdxx-2+1-2+1+ln 1+x+C-1x+ln 1+x+C

Page No 19.203:

Question 6:

2x+3x26x dx

Answer:

 2x+3x26xdx=2x2+3x2+2·2x·3x6xdx=2x22x·3x+3x22x·3x+2·2x·3x2x·3xdx23x+32x+2dx23xln 23+32xln 32+2x+C          axdx=axln a

Page No 19.203:

Question 7:

sin x1+sin x dx

Answer:

      sin x1+sin xdxRationalising the denominator sin x1+ sin x×1-sin x1-sin xdxsin x-sin2x1-sin2xdxsin xcos2x-tan2xdxsin xcos x×1cos x-sec2x-1dxsec x tan x-sec2x+1dxsec x-tan x+x+C

Page No 19.203:

Question 8:

x4+x2-1x2+1 dx

Answer:

x4+x 2-1x2+1dxx4+x2x2+1dx-1x2+1dxx2 x2+1x2+1dx-1x2+1dxx2 dx-1x2+1dxx33-tan-1 x+C              1x2+a2dx=1atan-1xa+C

Page No 19.203:

Question 9:

sec2 x cos2 2x dx

Answer:

sec2x·cos2 2xdx=sec2 x×2 cos2x-12dx=sec2x 4 cos4x-4 cos2x+1dx4 cos2x-4+sec2xdx=4cos2x dx+sec2x dx-4dx41+cos 2x2dx+sec2x-4dx2 x+sin 2x2+tan x-4x+Csin 2x+tan x-2x+C

Page No 19.203:

Question 10:

cosec2 x cos2 2x dx

Answer:

cosec2x·cos22x dxcosec2x 1-2 sin2x2dxcosec2x 1+4 sin4x-4 sin2x dxcosec2x+4 sin2x-4 dxcosec2x dx+41-cos 2x2dx-4dx-cot x+2 x-sin 2x2-4x+C-cot x+2x-sin 2x-4x+C-cot x-sin 2x -2x+C

Page No 19.203:

Question 11:

sin4 2x dx

Answer:

sin4 2x dxsin2 2x2 dx1-cos 4x22dx141-cos 4x2141+cos2 4x-2 cos 4xdx141+1+cos 8x2-2 cos 4xdx1432+cos 8x2-2 cos 4xdx143x2+sin 8x16-2 sin 4x4+C3x8+sin 8x64-sin 4x8+C

Page No 19.203:

Question 12:

cos3 3x dx

Answer:

cos3 3x dx                As we know that  cos 3A=4 cos3A-3 cos A  cos 3A+3 cos A4=cos3Acos 9x+3 cos 3x4dx14cos 9x dx+34cos 3x dx14 sin 9x9+34sin 3x3+Csin 9x36+sin 3x4+C1363 sin 3x-4 sin3 3x +sin 3x4+C        sin 3x=3 sin x-4 sin3xsin 3x12+sin 3x4-19 sin3 3x+Csin 3x3-sin3 3x9+C

Page No 19.203:

Question 13:

sin 2xa2+b2 sin2 x

Answer:

Let I=sin 2xa2+b2 sin2xdxPutting  a+b2 sin2x=tb2 2 sin x cos x dx=dtb2×sin 2x dx=dtI=1b2dtt     =1b2ln t+C        =1b2ln a2+b2 sin2x+C               t=  a+b2 sin2x

Page No 19.203:

Question 14:

1sin-1 x 1-x2 dx

Answer:

Let I=1sin-1x·1-x2dxPutting  sin-1 x=tdx1-x2=dt I=dtt       =ln t+C       =ln sin-1x+C       t=sin-1x

Page No 19.203:

Question 15:

sin-1 x31-x2 dx

Answer:

Let I=sin-1 x31-x2dxPutting  sin-1x=tdx1-x2=dt I=t3·dt      =t44+C      =sin-1 x44+C        t= sin-1x

Page No 19.203:

Question 16:

1ex+1 dx

Answer:

1ex+1dx       ...(1) 
Multiplying numerator and Denominator of eq (1) by ex

ex·dxex ex+1Putting  ex=tex dx=dtdtt t+11t t+1=At+Bt+11t t+1=A t+1+B tt t+1       ...(2)1=A t+1+ B tPutting t+1=0 or, t=-1 in eq (2) we get , 1=A×0+B -1B=-1Now,putting t=0 in eq (2) we get , 1=A 0+1+B×0A=1Putting the values of A and B in eq (2) we get , 1t t+1=1t-1t+1dtt t+1=dtt-dtt+1                  =ln t-ln t+1+C                  =ln tt+1+C                  =ln exex+1+C                  =ln ex-ln  ex+1+C                  =x-ln ex+1+C

Page No 19.203:

Question 17:

ex-1ex+1 dx

Answer:

We have,I=ex-1ex+1dx=2ex-ex+1ex+1dx=2exex+1dx-dxPutting ex+1=texdx=dtI=2tdt-dx=2 log t-x+C=2 log ex+1-x+C=2 log ex+1-x+C

Page No 19.203:

Question 18:

1ex+e-x dx

Answer:

Let I =1ex+e-xdx        =dxex+1ex        =ex dxe2x+1        =ex dxex2+1Putting ex=tex dx=dt I=dtt2+1      =tan-1t +C         dta2+x2=1atan-1xa+C      =tan-1 ex+C          t=ex

Page No 19.203:

Question 19:

cos7 xsin x dx

Answer:

Let I=cos7x sin xdx        =cos6x·cos x dxsin x        =cos2x3·cos xsin xdx        =1-sin2x3·cos xsin xdxLet sin x=tcos x dx =dt I=1-t23tdt      =1-t6-3t2+3t4tdt     =1t-t5-3t+3t3dt     =ln t-t66-3t22+3t44+C     =ln sin x-sin6x6-3 sin2 x2+34 sin4x+C         t=sin x

Page No 19.203:

Question 20:

sin x sin 2x sin 3x dx

Answer:

sin x·sin 2x·sin 3x dx=122 sin 2x·sin x sin 3x dx=12cos 2x-x- cos 2x+x sin 3x dx         2sinAsinB=cos (A-B)-cos (A+B)=12cos x-cos 3x sin 3x dx=12sin 3x·cos x dx-12sin 3x·cos 3x dx=142 sin 3x·cosx dx-142 sin 3x·cos 3x dx       =14sin 4x+sin 2xdx-14sin 6x dx                 2sin Acos B=sinA+B+sinA-B=14-cos 4x4-cos 2x2-14-cos 6x6+C=-cos 4x16-cos 2x8+cos 6x24+C

Page No 19.203:

Question 21:

cos x cos 2x cos 3x dx

Answer:

cos x·cos 2x·cos 3x dx122 cos 2x·cos x cos 3x dx12cos 2x+x+cos 2x-x cos 3x dx                               2cos Acos B=cos A+B+cosA-B12cos3x+cos x cos 3x dx12cos2 3x dx+12cos 3x·cos x dx121+cos 6x2dx+142 cos 3x·cos x dx                           cos 2x=  cos2 x-1 14x+sin 6x6+14cos 4x+cos 2xdx14x+sin 6x6+14sin 4x4+sin 2x2+Cx4+sin 6x24+sin 4x16+sin 2x8+C

Page No 19.203:

Question 22:

sin x+cos xsin 2x dx

Answer:

Let I=sin x+cos xsin 2 xdxPutting sin x-cos x=tcos x+sin xdx=dtAlso sin x-cos x2=t2sin2x+cos2x-2 sin x cos x=t21-t2=sin 2xI=dt1-t2    =sin-1 t+C                     dta2-x2=sin-1xa+C    =sin-1 sin x-cos x+C        t=sin x-cos x

Page No 19.203:

Question 23:

sin x-cos xsin 2x dx

Answer:

Let I=sin x-cos xsin 2x dxPutting  sin x+cos x=tcos x-sin x dx=dtsin x-cos x dx=-dtAlso sin x+cos x=tSquaring both sides, sin x+cos x2=t2sin2x+cos2x+2 sin x cos x =t21+sin 2x=t2sin 2x=t2-1I=-dtt2-1     =-ln t+t2-1+C                  dtx2-a2=lnx+x2-a2+C     =-ln sin x+cos x+sin x+cos x2-1+C            t=sin x+cos x     =-ln sin x+cos x+sin2x+cos2x+2 sin cos x-1+C     =-ln sin x+cos x+sin 2 x+C

Page No 19.203:

Question 24:

1sin x-a sin x-b dx

Answer:

1sin x-a·sin x-bdx=1sin b-asin b-asin x-a·sin x-b dx=1sin b-asin x-a-x-bsin x-a·sin x-b dx=1sin b-asin x-a·cos x-b-cos x-a sin x-bsin x-a·sin x-b dx=1sin b-a sin x-a·cos x-bsin x-a·sin x-b - cos x-a sin x-b sin x-a sin x-b dx=1sin b-acot x-b-cot x-a dx=1sin b-acot x-b dx-cot x-a dx=1sin b-aln sin x-b-ln sin x-a+C=1sin b-aln sin x-bsin x-a +C=-1sin a-blnsin x-bsin x-a +C=1sin a-b ln sin x-asin x-b+C

Page No 19.203:

Question 25:

1cos x-a cos x-b dx

Answer:

1cos x-a·cos x-bdx=1sin a-bsin a-bcos x-a·cos x-b dx=1sin a-bsin x-b-x-acos x-a·cos x-b dx=1sin a-bsin x-b·cos x-a-cos x-b·sin x-acos x-a·cos x-b=1sin a-bsin x-b·cos x-acos x-a·cos x-b-cos x-b·sin x-acos x-a·cos x-b dx=1sin a-btan x-b-tan x-a dx=1sin a-btan x-b dx-tan x-a dx=1sin a-bln sec x-b-ln sec x-a +C=1sin a-bln cos x-a-ln cos x-b +C=1sin a-bln cos x-acos x-b +C

Page No 19.203:

Question 26:

sin x1+sin x dx

Answer:

We have,I=sinx1+sinx dxI=2 sinx2cosx2sin2x2+cos2x2+2 sinx2cosx2 dxI=2 sinx2cosx2 sinx2+cosx22 dxI=2 sinx2cosx2 sinx2+cosx2 dxI=1+2sinx2 cosx2-1sinx+cosx dxI=sin2x2+cos2x2+2sinx2 cosx2-1sinx+cosx dxI=sinx2+cosx22-1sinx2+cosx2 dxI=sinx2+cosx22sinx2+cosx2 dx-1sinx2+cosx2 dxI=sinx2+cosx2 dx-1sinx2+cosx2 dxI=2-cosx2+sinx2+C1-12112sinx2+cosx2 dxI=2-cosx2+sinx2+C1-121sinx2 cosπ4+cosx2 sinπ4 dxI=2-cosx2+sinx2+C1-121sinx2+π4 dxI=2-cosx2+sinx2+C1-12cosecx2+π4 dxI=2-cosx2+sinx2-2logtanx4+π8+C

Page No 19.203:

Question 27:

sin xcos 2x dx

Answer:

Let I=sin xcos 2 x dx       =sin x2 cos2x-1 dx                          cos 2x=2cos2x-1                   Putting cos x=t-sin x dx=dtsin x dx=-dtI=-dt2t2-1     =12-dtt2-12     =-12dtt2-122     =-12×12×12 ln t-12t+12+C         1x2-a2=12alnx-ax+a+C     =-122 ln 2t-12t+1+C     =-122 ln 2 cosx-12 cos x+1+C             t= cos x     =122 ln 2 cos x+12 cos x-1+C

Page No 19.203:

Question 28:

tan3 x dx

Answer:

Let I =tan3 x dx=tan x·tan2x dx=tan x sec2x-1dx=tan x·sec2x dx-tan x dxPutting tan x=t in the Ist integralsec2x dx=dtI=t·dt-tan x dx      =t22-ln sec x+C      =tan2x2-ln sec x+C               t= tan x

Page No 19.203:

Question 29:

tan4 x dx

Answer:

Let I= tan4x dx       =tan2x·tan2x dx       =sec2x-1 tan2x dx       =sec2x·tan2x dx-tan2x dx       =tan2x·sec2x-sec2x-1 dxPutting  tan x=t in the Ist integralsec2x dx=dt I=t2·dt-sec2x-1 dx      =t33-tan x+x+C    =tan3x3-tan x+x+C              t=tan x

Page No 19.203:

Question 30:

tan5 x dx

Answer:

 Let I=tan5 x dx        =tan3x·tan2x dx        =tan3x sec2x-1 dx        =tan3x·sec2x dx-tan3x dx        =tan3x·sec2x dx-tan x·tan2x dx        =tan3x·sec2x dx-tan x·sec2x-1 dx        =tan3x·sec2x dx-tanx·sec2x dx+tan x dxPutting  tan x=t in the Ist and IInd integral. sec2x dx=dt I=t3·dt-t·dt+tan x dx      =t44-t22+ln sec x+C     =tan4x4-tan2x2+ln sec x+C              t= tan x

Page No 19.203:

Question 31:

cot4 x dx

Answer:

Let I=cot4x dx       =cot2x·cot2x dx      =cot2 x·cosec2x-1 dx      =cot2x·cosec2x dx-cot2x dx      =cot2x·cosec2x dx-cosec2x-1 dxPutting cot x=t in the Ist integral-cosec2x dx=dt I=-t2 dt-cosec2x-1 dx      =-t33+cot x+x+C      =-cot3x3+cot x+x+C              t= cot x

Page No 19.203:

Question 32:

cot5 x dx

Answer:

Let  I=cot5x dx         =cot2x·cot3x dx         =cosec2x-1 cot3x dx         =cot3x·cosec2x dx-cot3x dx         =cot3x·cosec2x dx-cot x·cot2x dx         =cot3x·cosec2x dx-cot x cosec2x-1 dx         =cot3x·cosec2x dx-cot x·cosec2x dx+cot x dxPutting  cot x=t in the Ist and IInd integral -cosec2x dx=dtcosec2x dx=-dtI=-t3 dt+t·dt+cot x dx     =-t44+t22+ln sin x+C     =-cot4x4+cot2x2+ln sin x+C              t=cot x

Page No 19.203:

Question 33:

x2x-13 dx

Answer:

x2x-13dx=x2-1+1x-13dx=x-1 x+1x-13+1x-13dx=x+1x-12+1x-13dx=x-1+2x-12+1x-13dx=1x-1+2x-12+1x-13dx=1x-1dx+2x-1-2 dx+x-1-3 dx=ln x-1+2 x-1-2+1-2+1+x-1-3+1-3+1+C=ln x-1 -2x-1-x-1-22+C=ln x-1 -2x-1-12 x-12+C

Page No 19.203:

Question 34:

x2x+3 dx

Answer:

Let I=x2x+3 dxPutting 2x+3=tx=t-322dx=dtdx=dt2 I=12t-32 t dt      =14t-3 t dt      =14t32-3t12dt      =14t32+132+1-3 t12+112+1+C      =14×25 t52-34×23 t32+C      =110 t52-2t32+C      =110 2x+352-12 2x+332+C          t=2x+3      =1102x+352-12 2x+332+C

Page No 19.203:

Question 35:

x31+x22 dx

Answer:

Let I=x31+x22dx       =x2×x1+x22dxPutting 1+x2=t x2=t-12x dx=dtx dx=dt2 I=12t-1t2dt      =121t-1t2dt      =12dtt-12t-2 dt      =12 ln t-12t-2+1-2+1+C      =12 ln t+12t+C      =12 ln 1+x2+12 1+x2+C          t= 1+x2

Page No 19.203:

Question 36:

x sin5 x2 cos x2 dx

Answer:

Let I=x·sin5 x2·cos x2 dxPutting sin x2=tcos x2×2x dx=dtcos x2·x dx=dt2 I=12t5·dt     =12 t66+C    =t612+C    =sin6 x212+C         t= sin x2

Page No 19.203:

Question 37:

sin3 x cos4 x dx

Answer:

Let I=sin3x·cos4x dx       =sin2x·sin x·cos4x dx       =1-cos2x·cos4x·sin x dx        =cos4x-cos6x·sin x dxPutting cos x=t-sin x dx=dtsin x dx=-dtI=-t4-t6dt     =t6-t4dt    =t77-t55+C    =cos7x7-cos5x5+C         t= cos x

Page No 19.203:

Question 38:

sin5 x dx

Answer:

Let I=sin5x dx       =sin4x·sin x dx       =sin2x2 sin x dx       =1-cos2x2 sin x dx       =cos4x-2 cos2x+1 sin x dxPutting cos x=t-sin x dx=dtsin x dx=-dt I=-t4-2t2+1 dt      =-t4 dt+2t2 dt-dt      =-t55+2t33-t+C      =-cos5x5+23 cos3x-cos x+C         t=cos x

Page No 19.203:

Question 39:

cos5 x dx

Answer:

Let I=cos5 x dx        =cos4x·cosx dx        =cos2x2 cos x dx         =1-sin2x2 cos x dxPutting sin x=tcos x dx=dt I=1-t22·dt      =t4-2t2+1 dt      =t4·dt-2t2 dt+dt      =t55-2×t2+12+1+t+C      =t55-23 t3+t+C      =sin5x5-23 sin3x+sin x+C       t= sin x

Page No 19.203:

Question 40:

sin xcos3 x dx

Answer:

Let I=sin x·cos3x dx        =sin x·cos2x·cos x dx        =sin x 1-sin2x·cos x dxPutting sin x=tcos x dx=dt I=t 1-t2·dt      =t12 dt-t12·t2 dt        =t12dt-t52dt        =t3232-t7272+C        =23t32-27 t72+C        =23 sin32x-27 sin72x+C              t= sin x

Page No 19.203:

Question 41:

sin 2xsin4 x+cos4 x dx

Answer:

Let I=sin 2xsin4x +cos4xdx        =2 sin x ·cos x dxsin4x+cos4xDividing numerator and denominator by cos4x2 sin x·cos xcos4 xdx1+tan4x2 tan x·sec2x dx1+tan2x2Putting tan2x=t2 tan x·sec2x dx I=dt1+t2      =tan-1 t+C      =tan-1 tan2x+C             t= tan 2x

Page No 19.203:

Question 42:

1x2-a2 dx

Answer:

Let I=1x2-a2dxPutting x= a sec θdx=a secθ tanθ dθ I=a secθ tanθ dθa2 sec2θ-a2       =a secθ·tanθ dθa·tanθ      =secθ dθ      =ln secθ+tanθ+C      =ln secθ+sec2θ-1+C      =ln xa+xa2-1+C      =ln x+x2-a2a+C      =ln x+x2-a2-ln a+C      =ln x+x2-a2+C'where C'=C-ln a

Page No 19.203:

Question 43:

1x2+a2 dx

Answer:

Let I=dxx2-a2Putting x= a tan θdx=a sec2θ dθ I=a·sec2θ dθa2 tan2θ+a2     =a sec2θ·dθa1+tan2θ     =sec2θ·dθsecθ     =secθ·dθ     =secθ·dθ     =ln secθ+tanθ+C     =ln secθ+sec2θ-1+C     =ln xa+x2a2-1+C     =ln x+x2-a2-ln a+C     =ln x+x2-a2+C'where C'=C-ln a

Page No 19.203:

Question 44:

14x2+4x+5 dx

Answer:

Let I=dx4x2+4x+1+4       =dx2x2+2×2x+1+22       =dx2x+12+22Putting 2x+1=t2 dx=dtdx=dt2 I=12dtt2+22      =12×12 tan-1 t2+C      =14 tan-1 2x+12+C              t=2x+1

Page No 19.203:

Question 45:

1x2+4x-5 dx

Answer:

1x2+4x-5dx=1x2+4x+4-4-5dx=1x2+4x+4-32dx=1x+22-32dx=12×3 ln x+2-3x+2+3+C          1x2-a2dx=12aln x-ax+a+C=16 ln x-1x+5+C

Page No 19.203:

Question 46:

11-x-4x2 dx

Answer:

We have,I=11-x-4x2dx=14114--x2x4dx=14114-x2+x4dx=14114-x2++182-182x4dx=14114-x+182+164dx=14114+-x+182164dx=14116+164-x+182dx=1411782-x+182dx=14×12×178 ln 178+x+18178-x-18+C          1a2-x2dx=12alna+xa-x+C=117 ln 17+18+x17-18-x+C

Page No 19.203:

Question 47:

13x2+13x-10 dx

Answer:

13x2+13x-10dx=131x2+133x-103dx=131x2+13 x3+1362-1362-103dx=131x+1362-16936-103dx=131x+1362-169-12036dx=131x+1362-1762dx=13×12×176 ln x+136-176x+136+176              1x2-a2dx=12aln x-ax+a+C=117 ln x-23x+5+C=117 ln 3x-23x+15+C

Page No 19.203:

Question 48:

sin xcos2 x-2 cos x-3 dx

Answer:

Let I=sin x cos2x-2 cos x-3dxPutting cos x=t-sin x dx=dtsin x dx=-dt I=-dtt2-2t-3      =-dtt2-2t+1-4      =-dtt-12-22      =-ln t-1+t-12-4+C                      1x2-a2dx=lnx+x2-a2+C      =-ln cos x-1+cos2x-2 cos x-3+C      t=cos x



Page No 19.204:

Question 49:

cosec x-1 dx

Answer:

Let I=cosec x-1 dx        =1sin x-1 dx        =1-sin xsin x dx       =1-sin x 1+sin xsin x 1+sin xdx       =1-sin2xsin2x+sin xdx       =cos xsin2x+sin xdxPutting sin x=tcos x dx=dtI=dtt2+t    =dtt2+t+122-122    =dtt+122-122    =ln t+12+t+122-122+C              1x2-a2dx=ln x+x2-a2+C    =ln t+12+t2+t+C    =ln sin x+12+sin2x+sin x+C              t= sin x

Page No 19.204:

Question 50:

13-2x-x2 dx

Answer:

 Let I=13-2x-x2dx        =13-x2+2x+1-1dx        =14-x+12dxPutting x+1=tdx=dt I=dt22-t2      =sin-1 t2+C                1a2-x2dx=sin-1xa+C      =sin-1 x+12+C           t = x+1

Page No 19.204:

Question 51:

x+1x2+4x+5 dx

Answer:

Let I=x+1x2+4x+5dx& let x+1= Addxx2+4x+5+Bx+1=A 2x+4+Bx+1=2Ax+4A+BEquating the coefficients of like terms2A=1A=12& 4A+B=14×12+B=1B=-1x+1=12 2x+4-1I=122x+4-1x2+4x+5dx     =122x+4x2+4x+5dx-1x2+4x+5dxPutting  x2+4x+5=t2x+4 dx=dt I=121tdt-1x2+4x+4+1dx      =12dtt-1x+22+12dx       =12 ln t-tan-1 x+21+C                     1x2+a2dx=1atan-1xa+C      =12 ln x2+4x+5-tan-1 x+2+C           t=x2+4x+5

Page No 19.204:

Question 52:

5x+7x-5 x-4 dx

Answer:

We have,I=5x+7x-5x-4 dx=5x+7x2-9x+20 dxLet 5x+7=A ddx x2-9x+20+B5x+7=A 2x-9+BEquating Coefficients of like terms2A=5A=52And-9A+B=7-9×52+B=7B=7+452B=592I=52 2x-9+592x2-9x+20 dx=522x-9 dxx2-9x+20+592dxx2-9x+20Putting x2-9x+20=t2x-9 dx=dtI=52dtt+592dxx2-9x+922-922+20=52t-12 dt+592dxx-922-81+804=52 t-12+1-12+1+592 dxx-922-122=52×2t+592 log x-92+x-922-122+C=5t+592 log x-92+x2-9x+20+C=5x2-9x+20+592 log x-92+x2-9x+20+C

Page No 19.204:

Question 53:

1+xx dx

Answer:

Let I=1+xxdx        =1+xx×1+x1+xdx        =1+xx2+xdxLet x+1= Addxx2+x+Bx+1=A 2x+1+Bx+1=2Ax+A+BEquating the coefficients of like terms2A=1A=12& A+B=112+B=1B=12 I=x+1x2+xdx      =12 2x+1+12x2+xdx      =122x+1x2+xdx+121x2+xdx

Putting x2+x=t2x+1 dx=dt I=121tdt+121x2+x+122-122dx      =121tdt+121x+122-122dx      =12t-12dt+121x+122-122dx      =12×2 t+12 ln x+12+x+122-14+C                    1x2-a2dx=ln x+x2-a2+C      =t+12 ln x+12+x2+x+C      =x2+x+12 ln x+12+x2+x+C              t=x2+x

Page No 19.204:

Question 54:

1-xx dx

Answer:

Let I=1-xxdx        =1-x·1-xx·1-x dx       =1-xx-x2dxLet 1-x= Addxx-x2+B1-x=A 1-2x+B1-x=-2A x+A+BEquating coefficients of like terms-2A=-1A=12& A+B=112+B=1B=12 I=12 1-2x+12x-x2dx       =121-2xx-x2dx+121x-x2+122-122dx      =121-2x x-x2dx+121122-x2-x+122dx      =121-2x x-x2dx+121122-x-122dx

Putting x-x2=t in the first integral1-2x dx=dt I=121tdt+121122-x-122dx      =12t-12dt+12dx122-x-122      =12×2 t12 +12×sin-1 x-1212+C                                   1a2-x2dx=sin-1xa+C      =t+12 sin-1 2x-1+C     =x-x2+12 sin-1 2x-1+C                                     t= x-x2

Page No 19.204:

Question 55:

a-x1-ax dx

Answer:

We have,I=a-x1-ax dxI=1a1+a-1-ax1-ax dxI=1a1-ax1-ax dx+1aa-11-ax dxI=1a dx+a-1a11-ax dxI=1ax+a-1a11-ax dxLet,I1=11-ax dxPut ax=z2adx=2zdzI1=1a2z1-z dzI1=1a2z-2+21-z dzI1=1a2z-21-z dz+1a21-z dzI1=-2a1-z1-z dz+1a21-z dzI1=-2a dz+1a21-z dzI1=-2az-2alog1-z+C1I1=-2axa-2alog1-ax+C1I=1ax+a-1a-2axa-2alog1-ax+C

Note: The answer in indefinite integration may vary depending on the integral constant.

Page No 19.204:

Question 56:

1sin x-2 cos x 2 sin x+cos x dx

Answer:

Let I=1sin x-2 cos x 2 sin x+cos xdx
Dividing numerator and denominator by cos2x we get ,
I= 1cos2xtan x-2 2 tan x+1dx =sec2xtan x-2 2 tan x+1 dxPutting tan x =tsec2x dx=dt I=1t-2 2t+1dt      =12t2+t-4t-2dt      =12t2-3t-2dt      = 121t2-3t2-1dt      =121t2-32t+342-342-1dt      =121t-342-916-1dt      =121t-342-542dt      =12×12×54 ln t-34-54t-34+54+C             1x2-a2dx=12alnx-ax+a+C      =15 ln t-2t+12+C      =15 ln 2 t-22t+1+C      =15 ln 2 tan x-22 tan x +1+C            t=tan x       =15 ln tan x-22 tan x+1+15 ln 2+C      =15 ln tan x-22 tan x+1+C'where C'=C+15 ln 2

Page No 19.204:

Question 57:

14 sin2 x+4 sin x cos x+5 cos2 x dx

Answer:

Let I=14 sin2x+4 sin x·cos x+5 cos2xdx

Dividing numerator and denominator by cos2x we get
I=sec2x 4 tan2x+4 tan x+5dxPutting tan x =tsec2x dx=dt I=dt4t2+4t+5      =14dtt2+t+54     =14dtt2+t+14-14+54     =14dtt+122+12     =14× tan-1 t+12+C             1x2+a2dx=1atan-1xa+C     =14 tan-1 2t+12+C     =14 tan-1 2 tan x+12+C           t= tan x     =14 tan-1 tan x+12+C    

Page No 19.204:

Question 58:

1a+b tan x dx

Answer:

Let I=1a+b tan xdx       =1a+b sin xcos xdx      =cos x·a cos x+b sin xdxLet cos x=A ddx a cos x+b sin x+B a cos x+b sin xcos x=A -a sin x+b cos x+B a cos x+ b sin x1·cos x=Ab+B·a cos x +sin x-A·a+B·bEquating coefficients of like terms   A·b+B·a=1          ...  1-A·a+B·b=0          ...  2Multiplying equation 1 by a and eq 2 by b and then adding them    A·ab+B·a2=a-A·a·b+Bb2=0B=aa2+b2Substituting the value of B in eq  1A·b+a2a2+b2=1A·b=1-a2a2+b2A=ba2+b2 I=ba2+b2-a sin x+b cos xa cos x+b sin xdx+aa2+b2a cos x+b sin xa cos x+b sin xdx     =ba2+b2-a sin x+b cos xa cos x +b sin xdx+aa2+b2dxPutting a cos x +b sin x =t in the Ist integral-a sin x+b cos xdx=dt I=ba2+b2dtt+aa2+b2dx     =ba2+b2 ln t+axa2+b2+C    =ba2+b2 ln a cos x+b sin x+axa2+b2+C         t=a cos x +b sin x 

Page No 19.204:

Question 59:

1sin2 x+sin 2x dx

Answer:

Let I=1sin2x+sin 2xdx       =1sin2x+2 sin x·cos xdx

Dividing numerator and denominator by cos2x, we get
I= 1cos2xtan2x+2 tan xdx =sec2xtan2x+2 tan x dxPutting tan x =tsec2x dx=dt I=1t2+2tdt     =1t2+2t+1-1dt     =1t+12-12dt     =12 ln t+1-1t+1+1+C     =12 ln tt+2+C      =12 ln tan xtan x+2+C                t= tan x

Page No 19.204:

Question 60:

sin x+2 cos x2 sin x+cos x dx

Answer:

sin x+2 cos x2 sin x+cos xdxLet sin x+2 cos x=A ddx 2 sin x+cos x+B 2 sin x+cos xsin x+2 cos x=A 2 cos x-sin x+2 B sin x+B cos xsin x+2 cos x=2 A+B cos x +2 B-A sin xEquating coefficients of like terms2 A+B=2          ...  1-A+2B=1          ...  2Multiplying eq 2 by 2 and adding it to eq 1 we get,5 B=4B=45Putting B=45 in eq 1 we get,2 A+45=2A=35sin x+2 cos x2 sin x+cos xdx= 35 2 cos x-sin x2 sin x+cos xdx+452 sin x+ cos x2 sin x+cos xdx                                            =352 cos x-sin x2 sin x+cos xdx+45dxPutting  2 sin x+cos x=t2 cos x-sin x dx=dt I=35dtt+45dx      =35 ln t+4x5+C      =35 ln 2 sin x+cos x+4x5+C           t= 2 sin x + cos x

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Question 61:

x3x8+4 dx

Answer:

Let I=x3x8+22dx      =x3x42+22dxPutting x4=t4x3dx=dtx3·dx=dt4 I=141t2+22dt      =14 ln t+t2+4+C      =14 ln x4+x8+4+C                          t= x4

Page No 19.204:

Question 62:

12-3 cos 2x dx

Answer:

Let I=12-3·cos 2xdx        =12-3 2 cos2x-1dx        =12-6 cos2x+3dx        =15-6 cos2xdx
Dividing numerator and denominator by cos2x, we get
I=sec2x 5 sec2x-6dx =sec2x 5 1+tan2x-6dx =sec2x 5 tan2x-1dxPutting tan x =tsec2x dx=dt I=15t2-1dt      =151t2-152dt     =15×12×15 ln t-15t+15+C        1x2-a2dx=12alnx-ax+a+C    =125 ln 5t-15t+1+C    =125 ln 5 tan x-15 tan x+1+C            t= tan x

Page No 19.204:

Question 63:

cos x14-cos2 x dx

Answer:

Let I=cos x 14-cos2xdx      =cos x 14-1-sin2xdx      =cos x sin2x-34dx      =cos x sin2x-322dxPutting sin x=tcos x dx=dt I=1t2-322dt     =12×32 ln t-32t+32+C     =13 ln 2t-32t+3+C     =13 ln 2 sin x-32 sin x+3+C                       t=sin x

Page No 19.204:

Question 64:

11+2 cos x dx

Answer:

Let I=11+2 cos xdx                 Putting cos x=1-tan2x21+tan2 x2 I=11+2 1-tan2 x21+tan2x2dx   =1+tan2 x2 1+tan2 x2+2-2 tan2 x2dx   =sec2 x2 3-tan2 x2dxPutting tanx2=t12sec2 x2 dx=dtsec2 x2·dx=2dt I=23-t2 dt     =2132-t2dt    =2×123 ln 3+t3+t+C        1a2-x2dx=12alna+xa-x+C   =13 ln 3+tanx23-tan x2+C           t= tan x2

Page No 19.204:

Question 65:

11-2 sin x dx

Answer:

Let I=11-2 sin x dx    Putting sin x=2 tan x21+tan2 x2 I=11-2 2 tan x21+tan2x2dx    =1+tan2 x21+tan2 x2-4 tan x2 dx    =sec2 x2 1+tan2 x2-4 tan x2dxPutting  tan x2=t12 sec2 x2 dx=dtsec2 x2·dx=2dt I=2 t2-4t+1dt    =21t2-4t+4-4+1dt    =21t-22-32dt    =2×123 ln t-2-3t-2+3+C            1x2-a2dx=12alnx-ax+a+C    =13 ln tanx2-2-3tanx2-2+3+C             t= tan x2

Page No 19.204:

Question 66:

1sin x 2+3 cos x dx

Answer:

Let  I=1sin x 2+3 cos xdx         =sin x sin2x 2+3 cos xdx         =sin x1-cos2x 2+3 cos x dx         =sin x1-cos x 1+cos x 2+3 cos x dxPutting cos x=t-sin x dx=dtI=-11-t 1+t 2+3tdt      =1t-1 t+1 3t+2dtLet1t-1 t+1 3t+2=At-1+Bt+1+C3t+21t-1 t+1 3t+2=A t+1 3t+2+B t-1 3t+2+C t+1 t-1t-1 t+1 3t+21=A t+1 3t+2+B t-1 3t+2+C t+1 t-1Putting  t+1=0 or t=-11=A×0+B -1 -1 3×-1+2+C×0B=12Now, putting t-1=0 or t=11=A 1+1 3+2+B×0+C×0A=110Now, putting  3t+2=0 or t=-2  31=A×0+B×0+C -23+1 -23-11=C 13 -53 C=-95 I=110 t-1dt+121t+1dt-9513t+2dt      =110 ln t-1+12 ln t+1-95 ln 3t+23+C      =110 ln t-1+12 log t+1-35 ln 3t+2+C      =110+ln cos x-1+12 ln cos x+1 -35 ln 3 cos x+2+C             t= cos x

Page No 19.204:

Question 67:

1sin x+sin 2x dx

Answer:

 Let  I=1sin x+sin 2xdx          =1sin x+2 sin x cos xdx          =1sin x 1+2 cos xdx          =sin x sin2x 1+2 cos xdx          =sin x 1-cos2x 1+2 cos xdx          =sin x dx1-cos x 1+cos x 1+2 cos xPutting  cos x=t-sin x dx=dtsin x dx=-dt

 I=-11-t 1+t 1+2tdt      =1t-1 t+1 2t+1dt1t-1 t+1 2t+1=At-1+Bt+1+C2t+11=A t+1 2t+1+B t-1 2t+1+C t-1 t+1Putting  t+1=0 or t=-11=A×0+B -1 -1 -2+1+C×01=B 2B=12Now, putting t-1=0 or t=11=A 2 3+B×0+C×0A=16Now, putting 2t+1=0 or t=-121=A×0+B×0+C -12-1 -12+11=C -32 12C=-43I =161t-1dt+121t+1dt-4312t+1dt      =16 ln t-1+12 log t+1-43 ln 2t+12+C      =16 ln t-1+12 ln t+1-23 ln 2t+1+C      =16ln cos x-1+12 ln cos x+1 -23 ln 2 cos x+1+C        t= cos x

Page No 19.204:

Question 68:

1sin4 x+cos4 x dx

Answer:

We have,I=dxsin4x+cos4x
Dividing numerator and denominator by cos4x
I=sec4x dxtan4x+1=sec2x sec2x dxtan4x+1=1+tan2x sec2x dxtan4x+1Putting tan x=tsec2x dx=dtI=1+t2 dtt4+1=1t2+1 dtt2+1t2=1+1t2t-1t2+2dtPutting t-1t=p1+1t2dt=dpI=1p2+22dp=12 tan-1p2+C=12 tan-1t-1t2+C=12 tan-1 t2-12 t+C=12 tan-1 tan2x-12 tan x+C=12 tan-1-2×1-tan2x2 tan x+C=12 tan-1-2tan 2x+C=12 tan-1-2 cot 2x+C

Page No 19.204:

Question 69:

15-4 sin x dx

Answer:

Let I=15-4 sin xdxPutting sin x=2 tan x21+tan2 x2 I=15-4 ×2 tan x21+tan2 x2dx       =1+tan2 x2 5 1+tan2 x2-8 tan x2dx       =sec2 x2 5 tan2 x2-8 tan x2+5dxPutting  tan x2=t12 sec2 x2 dx=dt

sec2 x2 dx=2 dt I=215t2-8t+5dt      =251t2-85t+1dt      =251t2-8t5+452-452+1dt      =251t-452-1625+1dt      =251t-452+352dt      =25×53 tan-1 t-4535+C          1x2+a2dx=1atan-1xa+C      =23 tan-1 5t-43+C      =23 tan-1 5 tan x2-43+C        t = tan x2

Page No 19.204:

Question 70:

sec4 x dx

Answer:

Let I= sec4x dx        =sec2x·sec2x dx       =1+tan2x·sec2x dxPutting  tan x=tsec2x dx=dtI=1+t2 dt     =dt+t2dt     =t+t33+C     =tan x+13tan3x+C        t= tan x 

Page No 19.204:

Question 71:

cosec4 2x dx

Answer:

Let I= cosec4 2x dx        =cosec2 2x·cosec2 2x dx        =1+cot2 2x·cosec2 2x dxPutting  cot 2x=t-cosec2 2x·2 dx=dtcosec2 2x·dx=-dt2 I=-121+t2·dt      =-12 t+t33+C      =-12cot 2x+16cot3 2x+C         t= cot 2x 

Page No 19.204:

Question 72:

1+sin xsin x 1+cos x dx

Answer:

Let I=1+sin x sin x 1+cos xdxPutting sin x=2 tan x21+ tan2 x2and cos x=1-tan2 x21+tan2 x2 I=1+2 tan x21+tan2 x2 2 tan x21+tan2 x2 1+1-tan2 x21+tan2 x2dx    =1+tan2 x2+2 tan x2 1+tan2 x2 2 tan x2 1+tan2 x2+1-tan2 x2dx    =141+tan2 x2+2 tan x2 sec2 x2tan x2 dxPutting  tan x2=t12 sec2 x2 dx=dtsec2 x2 dx=2dt I=141+t2+2t·2 dtt       =121t+t+2 dt       =12 ln t+t22+2t+C       =12 ln tan x2+tan2 x22+2 tan x2+C               t= tan x2        =12 ln tan x2+14tan2x2+tanx2+C

Page No 19.204:

Question 73:

12+cos x dx

Answer:

Let I=12+cos xdxPutting cos x =1-tan2 x21+tan2 x2 I=12+1-tan2 x21+tan2 x2dx     =1+tan2 x2 2 1+tan2 x2+1-tan2 x2dx     =sec2 x2 2+2 tan2 x2+1-tan2 x2dx     =sec2 x2 3+tan2 x2dxPutting  tan x2=t12 sec2 x2 dx=dtsec2 x2 dx=2 dt I=23+t2 dt      =21t2+32dt      =23 tan-1 t3+C       =23 tan-1 tan x23+C        t= tan x2

Page No 19.204:

Question 74:

a+xxdx

Answer:

Let  I=a+xxdx        =a+x a+xx a+x         =a+xx2+axdx        =a1x2+axdx+x x2+axdx        =a1x2+ax+a22-a22dx+x x2+axdx        =a1x+a22-a22dx+122x x2+axdx        =a1x+a22-a22dx+122x+a-ax2+axdx        =a1x+a22-a22dx+122x+a x2+axdx-a21x2+axdx        =a21x+a22-a22dx+122x+a x2+ax dxPutting  x2+ax=t in the Ist integral 2x+a dx=dt I=a21x+a22-a22dx+121tdt     =a2 ln x+a2+x2+ax+12×2t+C                 1x2-a2dx=lnx+x2-a2+C     =a2 ln x+a2+x2+ax+x2+ax+C                t=x2+ax

Page No 19.204:

Question 75:

6x+56+x-2x2 dx

Answer:

6x+5 dx6+x-2x2Let 6x+5= Addx6+x-2x2+B6x+5=A -4x+1+B6x+5=-4A x+A+BEquating coefficients of like terms-4A=6A=-32 & A+B=5-32+B=5B=5+32B=132Then, 6x+5=-32 -4x+1+1326x+5 6+x-2x2dx=-32-4x+1+1326+x-2x2dx                                      =-32-4x+16+x-2x2 dx+13216+x-2x2dxPutting 6+x-2x2=t in the Ist integral-4x+1 dx=dt 6x+5 6+x-2x2dx=-321tdt+132×213+x2-x2dx                                       =-32t-12·dt+132213+x2-x2-142+142dx                                      =-32t-12·dt+132213+116-x-142dx                                      =-32t-12·dt+13221742-x-142dx                                      =-3 t12+1322×sin-1 x-1474+C               1a2-x2dx=sin-1 xa+C                                      =-3 6+x-2x2+1322×sin-1 x-1474+C                                      =-36+x-2x2+1322 sin-1 4x-17+C

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Question 76:

sin5 xcos4 x dx

Answer:

Let I=sin5x cos4xdx       =sin4x.sin xcos4xdx       =sin2x2·sin x cos4xdx       =1-cos2x2.sin xcos4xdx       =1+cos4x-2cos2xcos4x sin x dxPutting cos x=t-sin x dx=dt I=-1+t4-2t2 dtt4      =-t-4 dt-dt+2t-2 dt      =-t-4+1-4+1-t+2 t-2+1-2+1+C      =13t3-t-2t+C      =13 cos3x-cos x-2cos x+C         t= cos x 

Page No 19.204:

Question 77:

cos5 xsin x dx

Answer:

Let I=cos5x dxsin x       =cos4x·cos x dxsin x      =cos2x2·cos x dxsin x      =1-sin2x2 cos x dxsin x      =1+sin4x-2 sin2xsin x cos x dxPutting sin x=tcos x dx=dt I=1+t4-2t2tdt      =dtt+t3 dt-2t dt      =ln t+t44-2t22+C      =ln t+t44-t2+C      =ln sin x+14sin4x-sin2x+C         t= sin x

Page No 19.204:

Question 78:

sin6 xcos x dx

Answer:

Let I=sin6x·cos xdx        =sin6x·cos xcos2xdx        =sin6x1-sin2xcos x dxPutting sin x=tcos x dx=dt I=t61-t2dt      =t6-1+11-t2 dt      =t23-131-t2dt+11-t2dt      =t2-1 1+t2+t41-t2+11-t2dt      =-t4+t2+1dt+11-t2dt      =-t55+t33+t+12 ln 1+t1-t+C      =-15sin5x-13sin3x-sin x+12 ln 1+sin x1-sin x+C            t= sin x

Page No 19.204:

Question 79:

sin2 xcos6 x dx

Answer:

Let I=sin2xcos6xdx       =sin2xcos2x·cos4xdx      =tan2x·sec4x dx      =tan2x sec2x·sec2x dx      =tan2x 1+tan2x sec2x dxPutting tan x=tsec2x dx=dt I=t2 1+t2dt      =t2+t4dt      =t33+t55+C      =13tan3x+15tan5x+C                        t= tan x

Page No 19.204:

Question 80:

sec6 x dx

Answer:

 Let I=sec6x dx        =sec4x·sec2x dx        =sec2x2·sec2x dx        =1+tan2x2 sec2x dxPutting tan x=tsec2x dx=dt I= 1+t22·dt        =1+t4+2t2dt        =dt+t4dt+2t2 dt        =t+t55+2t33+C        =tan x+15tan5 x+23 tan3x+C             t= tan x

Page No 19.204:

Question 81:

tan5 x sec3 x dx

Answer:

Let I=tan5x·sec3x dx        =tan4x·sec2x·sec x tan x dx        =sec2x-12·sec2x·sec x tan x dxPutting  sec x=tsec x tan x dx=dt I= t2-12·t2·dt       =t4-2t2+1t2 dt       =t6-2t4+t2 dt       =t77-2t55+t33+C       =17sec7x-25sec5x+13sec3x+C              t= sec x        

Page No 19.204:

Question 82:

tan3 x sec4 x dx

Answer:

Let I= tan3x·sec4x dx       =tan3x·sec2x·sec2x dx       =tan3x 1+tan2x·sec2x dx       =tan3x+tan5x sec2x dxPutting tan x=tsec2x dx=dt I= t3+t5 dt       =t44+t66+C       =tan4x4+ tan6x6+C         t= tan x

Page No 19.204:

Question 83:

1sec x+cosec x dx

Answer:

We have,I=1secx+cosecx dxI=11cosx+1sinx dxI=122sinx cosxsinx+cosx dxI=121+2sinx cosx-1sinx+cosx dxI=12sin2x+cos2x+2sinx cosx-1sinx+cosx dxI=12sinx+cosx2-1sinx+cosx dxI=12sinx+cosx2sinx+cosx dx-121sinx+cosx dxI=12sinx+cosx dx-121sinx+cosx dxI=12-cosx+sinx+C1-122112sinx+cosx dxI=12-cosx+sinx+C1-1221sinx cosπ4+cosx sinπ4 dxI=12-cosx+sinx+C1-1221sinx+π4 dxI=12-cosx+sinx+C1-122cosecx+π4 dxI=12-cosx+sinx-122logtanx2+π8+C

Page No 19.204:

Question 84:

a2+x2 dx

Answer:


Let I=1·IIa2+x2Idx       =a2+x2 1 dx-ddxa2+x2 1 dxdx       =a2+x2·x-1×2x2 a2+x2·x dx       =a2+x2·x-x2+a2-a2a2+x2dx       =xa2+x2-a2+x2 dx+a21a2+x2dx       =xa2+x2-I+a21a2+x2dx2I=xa2+x2+a2 ln x+x2+a2I=x2 a2+x2+a22 ln x+x2+a2+C

Page No 19.204:

Question 85:

x2-a2 dx

Answer:

Let I=1·IIx2-a2Idx       =x2-a21 dx-ddxx2-a21 dxdx       =x2-a2·x-1×2x2 x2-a2·x dx      =x2-a2·x-x2-a2+a2x2-a2dx      =x2-a2·x-x2-a2 dx-a2dxx2-a2      =xx2-a2-I-a2dxx2-a2 2I=xx2-a2-a2 ln x+x2-a2I=x2 x2-a2-a22 ln x+x2-a2+C

Page No 19.204:

Question 86:

a2-x2 dx

Answer:

Let I=a2-x2 dx      =1·IIa2-x2Idx      =a2-x21 dx-ddxa2-x21dxdx      =a2-x2·x+1×2x2 a2-x2·x dx      =a2-x2·x+x2-a2+a2a2-x2 dx      =xa2-x2-a2-x2 dx+a21a2-x2dx      =xa2-x2-I+a21a2-x2dx2I=xa2-x2+a2 1a2-x2dxI=x2 a2-x2+a22 sin-1 xa+C      

Page No 19.204:

Question 87:

3x2+4x+1 dx

Answer:

3x2+4x+1 dx=3x2+43x+13 dx=3x2+43x+232-232+13 dx=3x+232-49+13 dx=3x+232-132 dx=3 12x+23x+232-132 -12×132ln x+23+x+232-132 +C                    x2-a2 dx=12xx2-a2-12a2ln x+x2-a2+C=163x+23x2+4x+1-318ln  x+23+x2+43x+13 +C

Page No 19.204:

Question 88:

1+2x-3x2 dx

Answer:

1+2x-3x2dx=313+23x-x2dx=313-x2-23xdx=313-x2-23x+132-132dx=313+19-x-132dx=349-x-132dx=3232-x-132dx=3 x-132 232-x-132+2322 sin-1 x-1323+C               a2-x2dx=x2a2-x2+12a2sin-1xa+C=3x-16 1+2x-3x2+239sin-1 3x-12+C

Page No 19.204:

Question 89:

x1+x-x2 dx

Answer:

Let I=x1+x-x2dx& let x= Addx1+x-x2+Bx=A -2x+1+BBy equating the coefficients of like terms we get,x=-2A xA=-12& A+B=0B=12By substituting the values of A and B in eq (1) we get,I=-12 -2x+1+12 1+x-x2 dx =-12-2x+1 1+x-x2dx+12 1+x-x2dxPutting 1+x-x2=t-2x+1 dx=dt I=-12t·dt+121+x-x2 dx      =-12t dt+121-x2-x dx      =-12t12·dt+121-x2-x+14-14dx      =-12t12+112+1+121-x-122+14dx      =-12×23t32+12522-x-122dx      =-1 3 t32+12x-122 522-x-122+5222 sin-1 x-1252+C            a2-x2dx=x2a2-x2+12a2sin-1 xa+C      =-131+x-x232+122x-14 1+x-x2+58 sin-1 2x-15+C      =-1+x-x21+x-x23+2x-18 1+x-x2+516 sin-1 2x-15+C      =1+x-x2 -1+x-x23+2x-18+516 sin-1 2x-15+C      =1+x-x2 -8-8x+8x2+6x-324+516 sin-1 2x-15+C      =1+x-x2 8x2-2x-1124+516 sin-1 2x-15+C

Page No 19.204:

Question 90:

2x+3 4x2+5x+6 dx

Answer:

Let I=2x+3 4x2+5x+6 dx& let 2x+3=Addx4x2+5x+6+B2x+3=A 8x+5+B        ...(1)By equating coefficients of like terms we get,2x=8A xA=14& 5A+B=354+B=3B=3-54       =74Thus, by substituting the values of A and B in eq (1) we getI= 2x+3 4x2+5x+6 dx =148x+5+74 4x2+5x+6 dx =148x+5 4x2+5x+6 dx+74 4x2+5x+6 dxPutting 4x2+5x+6=t in the first integral8x+5 dx=dt I=14t·dt+7×24x2+5x4+32 dx      =14t12·dt+72x2-5x4+582-582+32dx      =14 t12+112+1+72x+582-2564+32dx      =14×23t32+72x+582+-25+9664      =16 t32+72x+582+7182      =1 6 4x2+5x+632+72x+582x+582+7182+7164×2 ln x+58+x+582+7182+C         a2+x2dx=12xa2+x2+12a2lnx+x2+a2+C       =16 4x2+5x+632+72 8x+516 x2+54x+32+71×72×128 ln  x+58+x2+54x+32+C      =16 4x2+5x+632+7×2 8x+54×16 x2+54x+32+497256 ln x+58+x2+54x+32+C      =16 4x2+5x+6 4x2+5x+6+764 8x+5 4x2+5x+6+497256 ln x+56+x2+54x+32 +C      =4x2+5x+6 4x2+5x+66+764 8x+5+497256 ln x+58+x2+54x+32+C      =4x2+5x+6 128x2+328x+297192+ln x+58+x2+54x+32+ C

Page No 19.204:

Question 91:

1+x2 cos 2x dx

Answer:

Let I=1+x2·cos 2x·dx       =cos 2x dx+x2·cos 2x dx       =sin 2x2+I1 where I1=x2 cos 2x dx      ...   1I1=x2I cos 2xII dx  = x2cos 2x dx-ddxx2 cos 2x dxdx  =x2·sin 2x2-2x×sin 2x2 dx  =x2·sin 2x2-xI·sin 2xII dx  =x2·sin 2x2-x sin 2x dx-ddxxsin 2x dxdx  =x2·sin 2x2-x-cos 2x2-1·-cos 2x2 dx  =x2·sin 2x2+x·cos 2x2-sin 2x4   ...   2From 1 and 2I=sin 2x2+x2 sin 2x2+x cos 2x2-sin 2x4+C     =x2+1 sin 2x2+x cos 2x2-sin 2x4+C

Page No 19.204:

Question 92:

log10 x dx

Answer:

log10 x dx=loge xloge 10  dx=1loge 101II·log xI dx=1loge 10loge x1 dx-ddxloge x1 dxdx=1loge 10loge x·x-1x×x dx=1loge 10x loge x-x+C=1loge10×x loge x-1+C=x loge x-1·log10e+C

Page No 19.204:

Question 93:

log log xx dx

Answer:

Let I=log log x dxxPutting log x=t1x dx=dt I=1II·log t·dIt       =log t1 dt-ddtlog t1 dtdt       =log t·t-1t×t dt       =log t·t-dt       =log t·t-t+C       =t log t-1+C       =log x log log x-1+C

Page No 19.204:

Question 94:

x sec2 2x dx

Answer:

xI·sec22x dxII=xsec22x dx-ddxxsec2 2x dxdx=x tan 2x2-1·tan 2x2 dx=x tan 2x2-12 ln sec 2x2+C=x tan 2x2-14 ln sec 2x+C

Page No 19.204:

Question 95:

x sin3 x dx

Answer:

x·sin3x dx                                         =x·143 sin x-sin 3x dx          sin3A-143 sin A-sin 3A=34xI·sin xII dx-14x·sin 3x dx=34xsin x dx-ddxxsin x dxdx-14xsin 3x dx-ddxxsin 3x dxdx=34x -cos x-1·-cos xdx-14x -cos 3x3-1·-cos 3x3dx=-3x4 cos x+34 sin x+x cos 3x12-sin 3x36+C=14-3x cos x+3sin x+x cos 3x3-sin 3x9+C

Page No 19.204:

Question 96:

x+12 ex dx

Answer:

x+12I exII dx=x+12ex dx-ddxx+12exdxdx=x+12·ex-2 x+1·ex dx=x+12 ex-2xI exII dx-2ex dx=x+12 ex-2 x·ex-1·ex dx-2ex=x+12 ex-2x ex+2ex-2ex+C=x+12-2x ex+C=x2+1 ex+C

Page No 19.204:

Question 97:

log x+x2+a2 dx

Answer:

Let I=1II·log x+x2+a2I dx       =log x+x2+a21 dx-ddxlog x+x2+a21dx       =log x+x2+a2·x-1x+x2+a2×1+1×2x2x2+a2·x·dx       =log x+x2+a2·x-xx2+a2dxPutting x2+a2=t in the second integral2x dx=dtx dx=dt2 I=x·log x+x2+a2-121tdt      =x·log x+x2+a2-12t-12dt      =x·log x+x2+a2-12 t-12+1-12+1+C      =x·log x+x2+a2-t+C      =x·log x+x2+a2-x2-a2+C        t=x2+a2

Page No 19.204:

Question 98:

log xx3 dx

Answer:

 log xx3dx=1x3II log xI dx=log x1x3dx-ddxlog x1x3dxdx=log xx-3 dx-1x×x-3+1-3+1dx=log x x-3+1-3+1+121x3dx=log x -12x2+12x-3 dx=log x -12x2+12 x-3+1-3+1+C=log x -12x2-14x2+C=-14x2 2 log x+1+C

Page No 19.204:

Question 99:

log 1-xx2 dx

Answer:

Let I=log 1-xx2dx      =1x2II log 1-xI dx      =log 1-xx-2 dx--11-x×x-2+1-2+1 dx      =log 1-x x-2+1-2+1+-11-x xdx      =log 1-x ×-1x+1x2-xdx      =-log 1-xx+1x2-x+122-122dx      =-log 1-x x+1x-122-122dx      =-log 1-xx+12×12 log x-12-12x-12+12+C      =-log 1-xx+log x-1x+C      =-log 1-xx+log x-1-log x+C      =-log 1-xx+log 1-x-log x+C      =1-1x log 1-x-log x+C

Page No 19.204:

Question 100:

x3 log x2 dx

Answer:

x3II·log x2I·dx=log x2x3 dx-2 log xx×x44 dx=log x2×x44-12log xI·x3II dx=log x2×x44-12log xx3 dx-ddxlog xx3dxdx=log x2×x44-12 log x·x44-1x×x44dx=log x2×x44-12 log x·x44-14x3 dx=log x2×x44-12 log x·x44-x416+C=log x2×x44-log x·x48+x432+C

Page No 19.204:

Question 101:

1x 1+xn dx

Answer:

We have,I=dxx 1+xn=xn-1 dxxn-1 x1 1+xn=xn-1 dxxn 1+xnPutting xn=tn xn-1 dx=dtxn-1 dx=dtnI=1ndtt 1+tlet 1+t=p2dt=2p dpI=1n2p dpp2-1 p=2ndpp2-12=2n×12 log p-1p+1+C=1n log 1+t-11+t+1+C=1n log 1+xn-11+xn+1+C

Page No 19.204:

Question 102:

x21-x dx

Answer:

We have,I=x21-x dxLet, 1-x=t2Differentiating both sides we get-dx=2t dtNow, integration becomes,I=-1-t22t 2tdt=-21-t22dt=-21-2t2+t4 dt=-2t-2t33+t55+C=-215t3t4-10t2+15+C=-2151-x31-x2-101-x+15+C=-2151-x31-2x+x2-101-x+15+C=-2151-x3x2-6x+3-10+10x+15+C=-2151-x3x2+4x+8+C

Page No 19.204:

Question 103:

x51+x3 dx

Answer:

We have,I=x5 dx1+x3=x3x2 dx1+x3Putting x3=t3x2 dx=dtx2 dx=dt3I=13t dt1+t=131+t-11+t dt=131+t-11+t dt=13 1+t3232-1+t-12+1-12+1+C=29 1+t32-23 1+t12+C=29 1+x332-23 1+x312+C=29 1+x3121+x3-3+C=29 1+x3 x3-2+C

Page No 19.204:

Question 104:

1+x21-x2 dx

Answer:

We have,I=1+x21-x2 dx=2-1-x21-x2 dx=211-x2 dx-1-x21-x2 dx=211-x2 dx-1-x2 dx=2 sin-1x-x21-x2+12sin-1x+C=2 sin-1x-x21-x2-12sin-1x+C=32 sin-1x-x21-x2+C

Page No 19.204:

Question 105:

x1-x1+x dx

Answer:

We have,I=x1-x1+x dxI=x1-x1-x1+x1-x dxI=x1-x1-x2 dxI=x-x21-x2 dxI=x-x2-1+11-x2 dxI=-x2+11-x2 dx+x-11-x2 dxI=1-x2 dx+x1-x2 dx-11-x2 dxI=x21-x2+12sin-1x+C1-1-x2+C2-sin-1x+C3          x1-x2 dx=-1-x2+C2I=1-x2x2-1-12sin-1x+C

Page No 19.204:

Question 106:

1x1+x3 dx

Answer:

We have,I=dxx 1+x3=x2dxx31+x3putting x3=t3x2 dx=dtx2dx=dt3I=13dtt1+tlet 1+t=p2dt=2p dpI=132p dpp2-1×p=23dpp2-1=23×12 log p-1p+1+C=13log p-1p+1+C=13log 1+t-11+t+1+C=13log  1+x3-11+x3+1+C

Page No 19.204:

Question 107:

sin x+cos xsin4 x+cos4 x dx

Answer:

We have,I=sin x+cos xsin4 x+cos4 x dx=sin x+cos xsin2 x+cos2 x2-2sin2 x cos2 x dx=sin x+cos x1-2sin2 x cos2 x dx=sin x+cos x1-122sin x cos x2 dx=sin x+cos x1-12sin22xdx

Putting sin x-cos x=t     .....1sin x-cos x2=t2sin2x+cos2x-2sin x cos x=t21-2sin x cos x=t2sin 2x=1-t2Differentiating 1, we getcos x+sin xdx=dtI=11-121-t22 dt=22-1-t22 dt=222-1-t22 dt=212+1-t22-1+t2 dt

=22212+1-t2+12-1+t2 dt=1212+1-t2 dt+1212-1+t2 dt=1212+12-t2 dt+1212-12+t2 dt=12×122+1log2+1+t2+1-t+12×12+1tan-1t2+1+C=12122+1log2+1+t2+1-t+12+1tan-1t2+1+C,   where t=sin x-cos x

Page No 19.204:

Question 108:

x2 tan-1 x dx

Answer:

We have,I= x2tan-1 x dxConsidering tan-1 x as first function and x2 as second functionI=tan-1xx33-11+x2×x33dx=tan-1xx33-13x3dx1+x2=tan-1 xx33-13x2x1+x2dxPutting 1+x2=tx2=t-12x dx=dtx dx=dt2I=tan-1xx33-16t-1tdt=x33tan-1x-16dt+16dtt=x33tan-1x-16t+16log t+C=x33tan-1x-161+x2+16log 1+x2+C=x33tan-1x-x26+16log x2+1+C'      Where C'=C-16



Page No 19.205:

Question 109:

tan-1x dx

Answer:

We have,I= tan-1 x dxPutting x=tan θx=tan2 θdx=2 tan θ sec2 θ dθI= tan-1tan θ 2tan θ sec2 θ dθ=2  θi tan θ sec2 θii dθ=2θ×tan2 θ2-1tan2 θ dθ2          tan θ sec2 θ dθ= tan2 θ2=2θtan2θ2-12sec2θ-1dθ=θ tan2 θ-2×tan θ2+2×θ2+C=tan-1x×x-x+tan-1 x+C=x+1tan-1 x -x+C

Page No 19.205:

Question 110:

sin-1x dx

Answer:

We have,I= sin-1 x dxPutting x=sin θx=sin2 θdx=2 sin θ cos θ dθdx=sin2θdθI= θ sin 2θdθ=θ-cos 2θ2-1-cos 2θ2dθ=-θ cos 2θ2+12cos 2θdθ=-θ cos 2θ 2+12sin 2θ2+C=-sin-1 x 1-2 sin2 θ2+122 sin θ cos θ2+C=-sin x1-2x2+sin θ1-sin2 θ2+C=-sin-1 x 1-2x2+x 1-x2+C=-12sin-1 x 1-2x+12x-x2+C

Page No 19.205:

Question 111:

sec-1x dx

Answer:

We have,I=sec-1 x dxPutting x=sec θx=sec2 θdx=2 sec θ sec θ tan θ dθ        =2 sec2 θ tan θ dθI=2θ sec2 θ tan θ dθ=2  θtan θ sec2 θ dθConsidering θ as first fucction and tan θ sec2 θ as second functionI=2θtan2 θ2-1tan2 θ2dθ          tan θ sec2 θ dθ=tan2 θ2=θ tan2 θ-sec2 θ-1dθ=θ tan2θ-tan θ+θ+C=θ1+tan2 θ-tan θ+C=θ sec2 θ-sec2 θ-1+C=sec-1x x-x-1+C=x sec-1x -x-1+C

Page No 19.205:

Question 112:

tan-11-x1+x dx

Answer:

We have,I= tan-11-x1+x dxPutting x=cos θdx=-sin θ dθI= tan-1 1-cos θ1+cos θ -sin θ dθ= tan-1 2 sin2 θ22 cos2 θ2 -sin θdθ= tan-1 tan θ2 -sin θdθ=-12θ sin θ dθConsidering θ as first function and sin θ as second functionI=-12θ-cos θ-1-cos θdθ=-12θ-cos θ+cos θdθ=-12-θ cos θ+sin θ+C=-12-θ cos θ+1-cos2 θ+C=-12-cos-1x ×x+1-x2+C=12x cos-1x-1-x2+C

Page No 19.205:

Question 113:

sin-1xa+x dx

Answer:

We have,I= sin-1 xa+x dxPutting x=a tan2 θtan θ=xadx=a2 tan θ sec2 θ dθI= sin-1 a tan2 θa+a tan2 θ 2a tan θ sec2 θdθ= sin-1 tan2 θsec2 θ 2a tan θ sec2 θ dθ=2a  sin-1 sin θ tan θ sec2 θ dθ=2a  θ tan θ sec2 θ dθConsidering θ as first function and tan θ sec2 θ as second functionI=2a θtan2θ2-1tan2 θ2dθ=aθ tan2 θ-sec2 θ-1dθ=aθ tan2 θ-tan θ+θ+C=aθ×1+tan2 θ-tan θ+C=atan-1xa 1+xa-xa+C=x+atan-1xa-ax+C

Page No 19.205:

Question 114:

sin-1 3x-4x3 dx

Answer:

We have,I= sin-1 3x-4x3dxPutting x=sin θθ=sin-1 xdx=cos θ dθI= sin-1 3 sin θ-4 sin3 θ cos θ dθ= sin-1 sin 3θ cos θ dθ=3 θI cos θII dθ=3 θ sin θ-1 sin θ dθ=3θ sin θ+cos θ+C=3θ sin θ+1-sin2 θ+C=3 sin-1 x ×x+1-x2+C=3 x sin-1x+1-x2+C

Page No 19.205:

Question 115:

sin-1 x3 dx

Answer:

We have,I=sin-1 x3 dxLet, sin-1x=tsin t=xcos t=1-x2Differentiating both sides we getcos t dt=dxNow, integral becomesI=sin-1 x3 dx=t3 cos t dt=t3sin t-3t2 sin t dt             Using by parts=t3sin t-3t2 sin t dt =t3sin t-3-t2 cos t --2t cos t dt =t3sin t+3t2 cos t-6t cos t dt =t3 sin t+3t2 cos t-6tsin t -sin t dt =t3 sin t+3t2 cos t-6t sin t+cos t +C=sin-1x3 x+3sin-1x2 1-x2-6sin-1x x-61-x2 +C=xsin-1xsin-1x2-6+3sin-1x2-21-x2 +C

Page No 19.205:

Question 116:

cos-1 1-2x2 dx

Answer:

We have,I= cos-1 1-2x2dxPutting x=sin θdx=cos θ dθI= cos-11-2 sin2θ cos θ dθ=cos-1 cos 2θ cos θ dθ=2 θI cos θII dθ=2θ sin θ-1 sin θ dθ=2θ sin θ+cos θ+C=2sin-1x×x+1-x2+C=2x sin-1x+1-x2+C

Page No 19.205:

Question 117:

x sin-1 x1-x23/2 dx

Answer:

We have,I=x sin-1x1-x232 dxPutting sin-1 x=θx=sinθdx=cosθ dθI=sinθ θ cosθ dθ1-sin2θ32=θ sinθ cosθ dθcos2θ32=θsinθcos2θ dθ=θ Isec θ tanθII dθ=θ×secθ-1×secθ dθ=θ×secθ-secθ dθ=θ×secθ-log secθ+tanθ+C=θcosθ-log 1cosθ+sinθcosθ+C=θ1-sin2θ-log 1+sinθcosθ+C=θ1-sin2θ-log 1+sinθ1-sin2θ+C=θ1-sin2θ-log 1+sinθ1-sinθ+C=sin-1x1-x2-log 1+x1-x+C=sin-1x1-x2-12 log 1+x1-x+C

Page No 19.205:

Question 118:

e2x 1+sin 2x1+cos 2x dx

Answer:

We have,I= e2x 1+sin 2x1+cos 2xdx= e2x 11+cos 2x+sin 2x1+cos 2xdx= e2x 12 cos2 x+2 sin x cos x2 cos2 xdx= e2xsec2 x2+tan xdxLet e2x tan x=te2xsec2x+2e2xtan xdx=dt=e2xsec2x2+e2xtan xdx=dt2I= dt2=t2+C=e2x tan x2+C

Page No 19.205:

Question 119:

1-sin x1+cos xe-x/2 dx

Answer:

We have,I=1-sin x1+cos xe-x2 dx=cos2 x2+sin2 x2-2 sin x2 cos x21+cos xe-x2dx=cos x2-sin x22e-x22 cos2 x2 dx=cos x2-sin x22 cos2 x2e-x2dx=12sec x2-tan x2 sec x2e-x2dxLet e-x2sec x2=te-x2 sec x2 tan x2×12-e-x2 sec x22 dx=dt12sec x2 tan x2-sec x2e-x2dx=dtI=-dt=-t+C=-e-x2sec x2+C

Page No 19.205:

Question 120:

ex1-x21+x22 dx

Answer:

We have,I=ex 1-x21+x22 dx=ex 1+x2-2x1+x22 dx=ex 1+x21+x22-2x1+x22 dx=ex 11+x2-2x1+x22 dx=ex1+x2+C                     exfx+f'x dx=exfx+Cwhere, fx=11+x2f'x=-2x1+x22

Page No 19.205:

Question 121:

em tan-1 x1+x23/2 dx

Answer:

We have,I=em tan-1 x1+x232dxPutting tan-1x=tx=tan t11+x2 dx=dtdx=1+x2dtdx=1+tan2tdtI=emt1+tan2 t321+tan2tdt=emtdt1+tan2 t=eIImtcosI t dt=cos temtm--sin temtm dt=cos temtm+1memtsin t dt=cos temtm+1m I1              .....1Where,I1=eIImtsin tI dt=sintemtm-costemtmdtI1=sin temtm-1mI                .....2from 1 and 2I=cos temtm+1m sin temtm-1mII=cos temtm+sin t emtm2-1m2 II+Im2=emt m cos t+sin tm2I=emt m cos t+sin t1+m2+CI=emt1+m2 cos tm1+m2+sin t11+m2+CLet m1+m2=cos θThen, sinθ=11+m2cotθ=mθ=cot-1mI=emt1+m2 cos t cos θ+sint sin θ+C=emt1+m2 cos t-θ+C=emt1+m2 cos tan-1x-cot-1m+C

Page No 19.205:

Question 122:

x2x-13 x+1 dx

Answer:

We have,I=x2x-13 x+1 dxLet x2x-13 x+1=Ax-1+Bx-12+Cx-13+Dx+1     .....1x2=Ax-12x+1+Bx-1x+1+C x+1+Dx-13     .....2Putting x=1 in 2, we get1=2CC=12Putting x=-1 in 2, we get1=-8DD=-18

Putting x=2 in 2, we get4=3A+3B+3C+D4=3A+3B+32-183A+3B=4-32+183A+3B=32-12+183A+3B=218A+B=78And putting x=0  in 2, we get0=A-B+C-D0=A-B+12+18           C=12, D=18A-B=-58
Here, A+B=78 and A-B=-58A=18 and B=34Therefore, 1 becomes,x2x-13 x+1=18x-1+34x-12+12x-13-18x+1Now, integral becomesI=18x-1+34x-12+12x-13-18x+1dx=18log x-1-34x-1-14x-12-18log x+1+C=18log x-1x+1-34x-1-14x-12+C

Page No 19.205:

Question 123:

xx3-1 dx

Answer:

We have,I=x dxx3-1=x dxx-1 x2+x+1Let xx-1 x2+x+1=Ax-1+Bx+Cx2+x+1xx-1 x2+x+1=A x2+x+1+Bx+C x-1x-1 x2+x+1x=A x2+x+1+Bx2-Bx+Cx-Cx=A+B x2+A-B+C x+A-CEquating Coefficient of like termsA+B=0                 .....1A-B+C=1           .....2A-C=0                 .....3Solving 1, 2 and 3, we getA=13B=-13C=13xx-1 x2+x+1=13 x-1+-13x+13x2+x+1=13 x-1+13 -x+1x2+x+1=13 x-1-13 x-1x2+x+1=13 x-1-16 2x-2x2+x+1=13 x-1-16 2x+1x2+x+1-16×-3x2+x+1=13 x-1-16 2x+1x2+x+1+12×1x2+x+1I=13dxx-1-162x+1 dxx2+x+1+12dxx2+x+14-14+1Putting x2+x+1=t2x+1 dx=dtI=13 log x-1-16 log t+12dxx+122+322=13 log x-1-16 log x2+x+1+1223 tan-1 x+1232+C=13 log x-1-16 log x2+x+1+13 tan-1 2x+13+C

Page No 19.205:

Question 124:

11+x+x2+x3 dx

Answer:

We have,I=dx1+x+x2+x3=dx1+x+x2 1+x= dx1+x 1+x2Let 1x+1 1+x2=Ax+1+Bx+Cx2+11x+1 x2+1=A x2+1+Bx+C x+1x+1 x2+11=A x2+1+Bx2+Bx+Cx+C1=A+B x2+B+C x+A+CEquating Coefficient of like termsA+B=0                 .....1B+C=0                 .....2A+C=1                 .....3Solving 1, 2 and 3, we get,A=12B=-12C=121x+1 x2+1=12 x+1+-x2+12x2+11x+1 x2+1=12 x+1-12 xx2+1+12 x2+1I=12dxx+1-12x dxx2+1+12dxx2+1Putting x2+1=t2x dx=dtx dx=dt2I=12dxx+1-14dtt+12dxx2+1=12 log x+1-14 log t+12 tan-1x+C=12 log x+1-14 log x2+1+12 tan-1 x+C=12 log x+1-12 log x2+1+12 tan-1 x+C=12 log x+1x2+1+12 tan-1 x+C

Page No 19.205:

Question 125:

1x2+2 x2+5 dx

Answer:

We have,I=dxx2+2 x2+5Putting x2=t1x2+2 x2+5=1t+2 t+5Let 1t+2 t+5=At+2+Bt+51t+2 t+5=A t+5+B t+2t+2 t+51=A t+5+B t+2Putting t=-51=B -5+2B=-13Putting t=-21=A -2+5+B×0A=13I=13dxx2+2-13dxx2+5=13dxx2+22-13dxx2+52=132 tan-1 x2-135 tan-1 x5+C

Page No 19.205:

Question 126:

x2-2x5-x dx

Answer:

We have,I=x2-2x5-x dx=x2-2x x4-1dx=x x2-2x2 x2-1 x2+1dxPutting  x2=t2x dx=dtx dx=dt2I=12t-2t t-1 t+1 dtLet t-2t t-1 t+1=At+Bt-1+Ct+1t-2t t-1 t+1=A t-1 t+1+Bt t+1+Ct·t-1t t-1 t+1t-2=A t-1 t+1+B t t+1+C t t-1Putting t=11-2=B×2B=-12Putting t=0-2=A -1A=2Putting t=-1-3=C -1 -2C=-32I=22dtt-12×2dtt-1-32×2dt+1=log t-14 log t-1-34 log t+1+C=log x2-14 log x2-1-34 log x2+1+C=2 log x-14 log x2-1-34 log x2+1+C

Page No 19.205:

Question 127:

1-x1+x dx

Answer:

We have,I=1-x1+x dxPutting x=cosθx=cos2θdx=-2 cosθ sinθ dθdx=-sin2θ dθI=1-cosθ1+cosθ -sin 2θ dθ=2 sin2 θ22 cos2 θ2 -2 sinθ cosθ dθ=sin θ2cos θ2 -2×2 sin θ2 cos θ2cosθ dθ=-4sin2 θ2×cosθ dθ=-41-cosθ2 cosθ dθ=-2cosθ-cos2θ dθ=-2cosθ-1+cos 2θ2dθ=-2cos θ dθ+1+cos 2θ dθ=-2sin θ+θ+sin 2θ2+C=-2 1-cos2θ+θ+2 sinθ cosθ2+C=-2 1-cos2θ+θ+sinθ cosθ+C=-21-x+cos-1x+1-xx+C=-21-x+cos-1x+x1-x+C

Page No 19.205:

Question 128:

x2+x+1x+12 x+2 dx

Answer:

We have,I=x2+x+1x+12x+2 dxLet x2+x+1x+12x+2=Ax+1+Bx+12+Cx+2     .....1x2+x+1=Ax+1x+2+Bx+2+Cx+12    .....2Putting x=-1 in 2, we get B=1Putting x=-2 in 2, we get C=3Putting x=0 in 2, we get1=2A+2B+C1=2A+2+3-4=2AA=-2Now, 1 becomesx2+x+1x+12x+2=-2x+1+1x+12+3x+2Therefore, integral becomesI=-2x+1+1x+12+3x+2dx=-2 log x+1-1x+1+3 log x+2+C

Page No 19.205:

Question 129:

sin 4x-21-cos 4xe2x dx

Answer:

We have,I=sin 4x-21-cos 4x e2x dx=2 sin 2x cos 2x-22 sin2 2x e2x dx=cot 2x-cosec2 2xe2x dxLet e2x cot 2x=t2e2xcot 2x+e2x-cosec2 2x×2 dx=dte2xcot 2x-cosec2 2x dx=dt2I=12dt=t2+C=12 e2xcot 2x+C

Page No 19.205:

Question 130:

cot x+cot3 x1+cot3 x dx

Answer:

We have,I=cot x+cot3x1+cot3x dx=cot x 1+cot2x1+cot3xdx=cot x cosec2x1+cot3x dxPutting cot x=t-cosec2x dx=dtcosec2x dx=-dtI=-t dt1+t3=-t dt1+t t2-t+1Let t1+t t2-t+1=At+1+Bt+Ct2-t+1t1+t t2-t+1=A t2-t+1+Bt+C t+1t+1 t2-t+1t=A t2-t+1+Bt2+Bt+Ct+Ct=A+B t2+B+C-A t+A+CEquating Coefficients of like termsA+B=0                 .....1B+C-A=1          .....2A+C=0               .....3Solving 1, 2 and 3, we getA=-13B=13C=13t1+t t2-t+1=-13 t+1+13 t+1t2-t+1t1+t t2-t+1=-13 t+1+16 2t+2t2-t+1t1+t t2-t+1=-13 t+1+16 2t-1+3t2-t+1I=--13dtt+1+162t-1t2-t+1 dt+12dtt2-t+1=+13dtt+1-162t-1t2-t+1 dt-12dtt2-t+14-14+1=13dtt+1-162t-1 dtt2-t+1-12dtt-122+322let t2-t+1=p2t-1 dt=dpI=13dtt+1-16dpp-12dtt-122+322=13 log t+1-16 log p-12×23 tan-1 t-1232+C=13 log t+1-16 log p-13 tan-1 2t-13+C=13 log cot x+1-16 log cot2x-cot x+1-13 tan-1 2 cot x-13+C



Page No 19.23:

Question 1:

2x-35+3x+2 dx

Answer:

      2x-35+3x+2dx=2x-35dx+3x+212dx=2x-35+125+1+3x+212+1312+1+C=2x-3612+293x+232+C

Page No 19.23:

Question 2:

17x-53+15x-4 dx

Answer:

      17x-53+15x-4dx=7x-5-3+5x-4-12dx=7x-5-3+17-3+1+5x-4-12+15-12+1+C=7x-5-2-14+255x-412+C

Page No 19.23:

Question 3:

12-3x+13x-2 dx

Answer:

      12-3x+13x-2dx=dx2-3x+3x-2-12dx=ln 2-3x-3+3x-2-12+13-12+1+C=ln 2-3x-3+233x-212+C=-13ln 2-3x+233x-2+C

Page No 19.23:

Question 4:

x+3x+14 dx

Answer:

      x+3x+14dx=x+1+2x+14dx=x+1x+14+2x+14dx=dxx+13+2dxx+14=x+1-3 dx+2x+1-4dx=x+1-3+1-3+1+2x+1-4+1-4+1+C=-12x+1-2-23x+1-3+C=-12x+12-23x+13+C

Page No 19.23:

Question 5:

1x+1+x dx

Answer:

dxx+1+x

Rationalise the denominator

=x+1-xx+1+xx+1-xdx=x+1-xx+1-xdx=x+112dx-x12dx=x+112+112+1-x12+112+1=23x+132-23x32+C

Page No 19.23:

Question 6:

12x+3+2x-3 dx

Answer:

dx2x+3+2x-3
Rationalise the denominator
=2x+3-2x-32x+3+2x-32x+3-2x-3dx=2x+3-2x-32x+3-2x-3dx=162x+312dx-162x-312dx=162x+312+1212+1-162x-312+1212+1+C=1182x+332-2x-332+C

Page No 19.23:

Question 7:

2x2x+12 dx

Answer:

     2x2x+12dx=2x+1-12x+12dx=2x+12x+12-12x+12dx=dx2x+1-2x+1-2dx=log2x+12-2x+1-2+12-2+1+C=log 2x+12+2x+1-12+C=log 2x+12+122x+1+C

Page No 19.23:

Question 8:

1x+a+x+b dx

Answer:

        dxx+a+x+b=x+a-x-bx+a+x+bx+a-x+bdx=x+a-x+bx+a-x+bdx=1a-bx+a12-1a-bx+b12dx=1a-bx+a12+112+1-1a-bx+b12+112+1+C=23a-bx+a32-x+b32+C

Page No 19.23:

Question 9:

sin x1+cos 2x dx

Answer:

      sin x 1+cos 2x dx=sin x.2 cos2x dx                    1+cos 2x=2 cos2x=2sin x cos x dx=222 sin x cos x dx=12sin 2xdx=12-cos 2x2+C=-122cos 2x+C

Page No 19.23:

Question 10:

1+cos x1-cos x dx

Answer:

      1+cos x1-cos x dx=2 cos2x22 sin2x2 dx       1+cos x=2 cos2x2  & 1-cos x=2 sin2x2 =cot2x2 dx=cosec2 x2-1 dx=-cot x212-x+C=-2 cot x2-x+C

Page No 19.23:

Question 11:

1-cos x1+cos x dx

Answer:

      1-cos x1+cos x dx=2 sin2x22 cos2x2 dx     1-cos x=2 sin2x2  & 1+cos x=2 cos2x2=tan2x2 dx=sec2 x2-1 dx=tan x212-x+C=2 tan x2-x+C

Page No 19.23:

Question 12:

11-sinx2 dx

Answer:

       dx1-sinx2=1+sin x21-sin x2 1+sin x2 dx=1+sin x21-sin2 x2dx=1+sinx2cos2x2 dx=sec2 x2+sec x2 tan x2dx=tan x212+sec x212+C=2 tan x2+sec x2+C

Page No 19.23:

Question 13:

11+cos 3x dx

Answer:

      dx1+cos 3x=1-cos 3x1+cos 3x 1-cos 3xdx=1-cos 3x1-cos2 3x dx=1-cos 3xsin2 3x dx=cosec2 3x dx-cosec 3x cot 3xdx=-cot 3x3+cosec 3x3+C=13cosec 3x-cot 3x+C=131sin 3x-cos 3xsin 3x+C=13 1-cos 3xsin 3x+C

Page No 19.23:

Question 14:

ex+12 ex dx

Answer:

      ex+12ex dx=e2x+2ex+1ex dx=e3x+2e2x+ex dx= e3x3+2e2x2+ex+C

Page No 19.23:

Question 15:

ex+1ex2 dx

Answer:

      ex+1ex2 dx=e2x+1e2x+2ex×1ex dx=e2x+e-2x+2dx=e2x2+e-2x-2+2x+C=e2x2-e-2x2+2x+C

Page No 19.23:

Question 16:

1+cos 4xcot x-tan x dx

Answer:

      1+cos 4xcot x-tan x dx=1+cos 4xcos xsin x-sin xcos x dx=2 cos2 2x×sin x cos xcos2x-sin2xdx=cos2 2x×2 sin x cos xcos 2xdx=cos 2x sin 2xdx=122 sin 2x cos 2xdx=12sin 4xdx=12-cos 4x4+C=-18cos 4x+C

Page No 19.23:

Question 17:

1x+3-x+2 dx

Answer:

dxx+3-x+2
Rationalising the denominator
=x+3+x+2x+3-x+2 x+3+x+2 dx=x+312+x+212x+3-x+2dx=x+312+x+212dx=x+312+112+1+x+212+112+1+C=23x+332+23x+232+C=23x+332+x+232+C

Page No 19.23:

Question 18:

tan2 2x-3 dx

Answer:

      tan2 2x-3dx=sec2 2x-3-1dx=sec2 2x-3dx-1dx=tan 2x-32-x+C



Page No 19.24:

Question 19:

1cos2 x 1-tan x2 dx

Answer:

Let I= 1cos2x 1-tan x2dx=sec2x1-tan x2dx=sec2x dx1-tan x2

Let 1 - tan x = t
-sec2x dx=dtsec2x dx=-dt

 I=-dtt2      =-t-2 dt      =-t-2+1-2+1+C      =1t+C       =11-tan x+C



Page No 19.30:

Question 1:

x2+5x+2x+2 dx

Answer:

      x2+5x+2x+2dx=x2x+2dx+5x dxx+2+2dxx+2=x2-4+4x+2dx+5x+2-2x+2dx+2dxx+2=x-2x+2x+2dx+4x+2dx+51-2x+2dx+2dxx+2=x-2 dx+4dxx+2+5dx-10dxx+2+2dxx+2=x-2dx-4dxx+2+5dx=x22-2x-4 ln x+2+5x+C=x22+3x-4 ln x+2+C

Page No 19.30:

Question 2:

x3x-2 dx

Answer:

      x3x-2dx=x3-8+8x-2dx=x3-23x-2+8x-2dx=x-2x2+2x+4x-2+8x-2dx=x2+2x+4dx+8dxx-2=x33+2x22+4x+8 ln x-2+C=x33+x2+4x+8 ln x-2+C

Page No 19.30:

Question 3:

x2+x+53x+2 dx

Answer:

      x2+x+53x+2dx=199x2+9x+453x+2dx=199x2-43x+2dx+9x+63x+2dx+433x+2dx=193x-23x+23x+2dx+33x+23x+2dx+43dx3x+2=193x-2 dx+31dx+43dx3x+2=193x22-2x+3x+43 3ln 3x+2+C=1932x2+x-433 ln 3x+2+C=16x2+19x-4327 ln 3x+2+C

Page No 19.30:

Question 4:

2x+3x-12 dx

Answer:

     2x+3x-12dx=2x-2+2+3x-12dx=2x-1+5x-12dx=2dxx-1+5x-1-2 dx=2 ln x-1+5x-1-2+1-2+1+C=2 ln x-1-5x-1+C

Page No 19.30:

Question 5:

x2+3x-1x+12 dx

Answer:

Let I= x2+3x-1x+12dx
Putting x + 1 = t
x = t – 1
& dx = dt

 I=t-12+3 t-1-1t2dt     = t2-2t+1+3t-3-1t2dt     =t2+t-3t2dt     =1+1t-3t-2dt     =t+log t-3t-2+1-2+1+C     =t+log t+3t+C     =x+1+log x+1+3x+1+C          t=x+1
Let C + 1 = C′
=x+log x+1+3x+1+C

Page No 19.30:

Question 6:

2x-1x-12 dx

Answer:

      2x-1x-12dx=2x-2+2-1x-12dx=2 x-1x-12+1x-12dx=2dxx-1+x-1-2 dx=2 ln x-1+x-1-2+1-2+1+C=2 ln x-1-1x-1+C



Page No 19.33:

Question 1:

x+12x+3 dx

Answer:


x+12x+3dx=122x+22x+3dx=122x+3-12x+3dx=122x+32x+3-12x+3dx=122x+3-12x+3dx=122x+312dx-2x+3-12dx=122x+312+1212+1-2x+3-12+12-12+1+C=12132x+332-2x+312+C=162x+332-122x+312+C

Page No 19.33:

Question 2:

xx+2 dx

Answer:

Let I=xx+2dx
Putting  x + 2 = t
Then, x = t – 2
Difference both sides
dx = dt
Now, integral becomes
I=t-2tdt=t32-2t12dt=t32+132+1-2t12+112+1+C=25t52-43t32+C=25x+252-43x+223+C

Page No 19.33:

Question 3:

x-1x+4 dx

Answer:

Let I=x-1x+4dx

Putting  x + 4 = t
Then, x = t – 4
Difference both sides
dx = dt
Now integral becomes,
I= t-4-1tdt=tt-5tdt=t12-5t-12dt=t12+112+1-5t-12+1-12+1+C=23t32-10t+C=23x+432-10x+412+C

Page No 19.33:

Question 4:

x+2 3x+5 dx

Answer:

Let I=x+2 3x+5 dxPutting 3x+5=tx=t-53

3dx=dtdx=dt3

 I=t-53+2tdt3      =13t-5+63t dt      =19t+1 t dt      =19t32+t12 dt      =19t32+132+1+t12+112+1+C      =1925t52+23t32+C      =19253x+552+233x+532+C         t=3x+5      =293x+5323x+55+13+C      =293x+5329x+15+515+C      =293x+5329x+2015+C      =21353x+5329x+20+C

Page No 19.33:

Question 5:

2x+13x+2 dx

Answer:

      2x+13x+2dx=136x+33x+2dx=136x+4-13x+2dx=1323x+23x+2-13x+2dx=1323x+2-13x+2dx=1323x+2 12dx-3x+2 -12dx=1323x+212+13 12+1-3x+2-12+1-12+1×3+C=13493x+232-233x+212+C=4273x+232-293x+212+C=3x+24273x+2-29+C=3x+243x+2-627+C=3x+212x+8-627+C=2276x+13x+2+C

Page No 19.33:

Question 6:

3x+57x+9 dx

Answer:

Let I=3x+57x+9dxPutting 7x+9=tx=t-97

& 7dx=dtdx=dt7

 I=3t-97+5tdt       =37tt-277t+5tdt7       =37×7t12dt-277×7t-12dt+57t-12dt       =37×7t12+112+1-277×7t-12+1-12+1+57t-12+1-12+1+C       =27×7t32-277×7×2 t12+10t7+C       =27×77x+932-547×77x+912+1077x+9+C         t=7x+9       =27×77x+932+10-547 7x+97+C       =277x+932+70-547 7x+97+C       =27×77x+932+167×77x+9+C       =27×77x+9127x+9+8+C       =2497x+9127x+17+C

Page No 19.33:

Question 7:

xx+4 dx

Answer:

xx+4dx=x+4-4x+4dx=x+4-4x+4dx=x+412dx-4x+4-12dx=x+412+112+1-4x+4-12+1-12+1+C=23x+432-8 x+412+C=x+41223x+4-8+C=x+4122x+8-243+C=x+4122x-163+C=23x-8x+4+C

Page No 19.33:

Question 8:

2-3x1+3x dx

Answer:

Let I=2-3x1+3xdx
Putting 1 + 3x = t
⇒ 3x = t – 1

& 3dx=dtdx=dt3

 I=2-t-1tdt      =3-ttdt      =3t-12-t12dt      =3t-12dt-t12dt      =3t-12+1-12+1-t12+112+1+C      =6t-23t32+C      =2t 3-t3+C      =2t9-t3+C          t=1+3x      =231+3x 9-1+3x3+C      =23×31+3x 8-3x+C      =298-3x 1+3x+C

Page No 19.33:

Question 9:

5x+3 2x-1 dx

Answer:

Let I=5x+3 2x-1dxPutting 2x-1=t2x=t+1x=t+12

& 2dx=dtdx=dt2

I=5t+12+3·t·dt2     =5 t2+52+3×t dt2     =145t+11 t12 dt     =145t32+11t12dt     =145t32+132+1+11t12+112+1+C     =14×25×5 t52+14×11×23 t32+C     =12t52+116t32 +C     =t322t+113+C     =t3223t+113 +C     =2x-132232x-1+113+C              t=2x-1     =2x-13226x-3+113+C     =2x-12322 3x+43+C     =2x-1323x+43+C

Page No 19.33:

Question 10:

xx+a-x+bdx

Answer:

xx+a-x+bdx=xx+a-x+b×x+a+x+bx+a+x+bdx=xx+a+x+bx+a2-x+b2dx=xx+a+x+bx+a-x-bdx=1a-bxx+a+x+b dx=1a-bxx+a dx+xx+b dx=1a-bx+a-ax+a dx+x+b-bx+b dx=1a-bx+ax+a dx-ax+a dx+x+bx+b dx-bx+b dx=1a-bx+a32dx-ax+a12dx+x+b32dx-bx+b12dx=1a-bx+a5252-ax+a3232+x+b5252-bx+b3232+c        where, c is an arbitrary constant=1a-b25x+a52-2a3x+a32+25x+b52-2b3x+b32+c        where, c is an arbitrary constantHence, xx+a-x+bdx=1a-b25x+a52-2a3x+a32+25x+b52-2b3x+b32+c        where, c is an arbitrary constant



Page No 19.36:

Question 1:

sin2 2x+5 dx

Answer:

     sin2 2x+5dx=1-cos 4x+102dx            sin2A=1-cos2A2=121-cos 4x+10dx=12x-sin 4x+104+C=12x-sin 4x+108+C

Page No 19.36:

Question 2:

sin3 2x+1 dx

Answer:

sin3 2x+1dx=143 sin 2x+1-sin 32x+1dx          sin 3θ=3 sinθ-4sin3θsin3θ=143sin θ-sin 3θ =34sin 2x+1dx-14sin 6x+3dx=34-cos 2x+12-14-cos 6x+36+C=-3   8cos 2x+1+124 cos 6x+3+C

Page No 19.36:

Question 3:

cos4 2x dx

Answer:

     cos4 2x dx=cos2 2x2dx=1+cos 4x22dx                   cos2 x=1+cos 2x2=141+cos 4x2 dx=141+cos2 4x+2 cos 4xdx=141+1+cos 8x2+2 cos 4xdx=1432+cos 8x2+2 cos 4xdx=143x2+sin 8x16+2 sin 4x4+C=3x8+sin 8x64+sin 4x8+C

Page No 19.36:

Question 4:

sin2 b x dx

Answer:

     sin2 bx dx=1-cos 2bx2dx           sin2 x=1-cos 2x2=121-cos 2bxdx=12x-sin 2bx2b +C=x2-sin 2bx4b+C

Page No 19.36:

Question 5:

sin2x2 dx

Answer:

     sin2 x2 dx=1-cos x2dx      sin2 x2=1-cos x2=121-cos xdx=12x-sin x +C

Page No 19.36:

Question 6:

cos2x2 dx

Answer:

     cos2 x2 dx=1+cos x2dx                 cos2 x2=1+cos x2=121+cos xdx=12x+sin x +C

Page No 19.36:

Question 7:

cos2 nx dx

Answer:

      cos2nx dx=1+cos 2nx2 dx          cos2x=1+cos 2x2=121+cos 2nxdx=12x+sin 2nx2n+C=x2+sin 2nx4n+C

Page No 19.36:

Question 8:

sin x 1-cos 2x dx

Answer:

      sin x·1-cos 2x dx=sin x .2 sin2x dx            1-cos 2A=2sin2A=2sin2x dx=21-cos 2x2dx=121-cos 2xdx=12x-sin 2x2+C



Page No 19.38:

Question 1:

sin 4x cos 7x dx

Answer:

sin 4x cos 7x dx=122 cos 7x sin 4x dx=12sin 7x+4x-sin 7x-4xdx          2 cos A sin B = sin A+B- sin A-B=12sin 11x-sin 3x dx=12-cos 11x11+cos 3x3+C=-cos 11x22+cos 3x6

Page No 19.38:

Question 2:

cos 3x cos 4x dx

Answer:

cos 4x cos 3x dx=122 cos 4x cos 3x dx=12cos 4x+3x+cos 4x-3xdx    2 cos A cos B=cos A+B+cos A-B=12cos 7x+cos x dx=12sin 7x7+sin x+C=114sin 7x+12sin x+C

Page No 19.38:

Question 3:

cos mx cos nx dx mn

Answer:

      cos mx cos nx dx=122 cos mx cos nxdx=12cos mx+nx+cos mx-nxdx         2 cos A cos B=cos A+B+cos A-B=12sin m+nxm+n+sin m-nxm-n+C

Page No 19.38:

Question 4:

sin mx cos nx dx mn

Answer:

      sin mx·cos nx dx=122 sin mx·cos nxdx=12sin mx+nx+sin mx-nxdx      2 sin A·cos B=sin A+B+sin A-B=12-cos m+nxm+n-cos m-nxm-n+C

Page No 19.38:

Question 5:

Integrate the following integrals:

sin2x sin4x sin6x dx

Answer:

​sin2x sin4x sin6x dx=122 sin2x sin4x sin6x dx=12cos2x-4x-cos2x+4x sin6x dx=12cos2x-cos6x sin6x dx=12cos2xsin6x dx-cos6xsin6x dx=142cos2xsin6x dx-2cos6xsin6x dx=14sin2x+6x-sin2x-6x dx-sin12x dx=14sin8x dx+sin4x dx-sin12x dx=14-cos8x8 +-cos4x4+cos12x12+c=-cos8x32 -cos4x16+cos12x48+c

Hence, sin2x sin4x sin6x dx=-cos8x32 -cos4x16+cos12x48+c.

Page No 19.38:

Question 6:

Integrate the following integrals:

sinx cos2x sin3x dx

Answer:

​sinx cos2x sin3x dx=122 sinx cos2x sin3x dx=12sinx+2x+sinx-2x sin3x dx=12sin3x-sinx sin3x dx=12sin23x dx-sinxsin3x dx=142sin23x dx-2sinxsin3x dx=141-cos6x dx-cosx-3x-cosx+3x dx=141 dx-cos6x dx-cos2x dx+cos4x dx=14x-sin6x6 -sin2x2+sin4x4+c=x4-sin6x24 -sin2x8+sin4x16+c

Hence, sinx cos2x sin3x dx=x4-sin6x24 -sin2x8+sin4x16+c.



Page No 19.4:

Question 1:

Evaluate each of the following integrals:

(i) x4dx

(ii) x54dx

(iii) 1x5dx

(iv) 1x3/2dx

(v) 3xdx

(vi) 1x23dx

(vii) 32 log3 xdx

(viii) logxxdx

Answer:

(i)
x4dx=x4+14+1+C =x55+C

(ii)
x54dx=x54+154+1+C=49x94+C

(iii)
x-5dx=x-5+1-5+1+C=-14x-4+C=-14x4+C

(iv)
dxx3/2=x-3/2dx=x-32+1-32+1+C=x-12-12+C=-2x+C

 (v)
3xdx=3xln 3+C

(vi)
     dxx23=dxx2/3=x-2/3 dx=x-23+1-23+1+C=3x13+C

(vii)
    32 log3xdx=3log3 x2dx=x2dx=x33+C

(viii)
  logxx dx=1·dx=x+C

Page No 19.4:

Question 2:

Evaluate: (i) 1+cos 2x2dx
(ii) 1-cos 2x2dx

Answer:

(i)
    1+cos 2x2dx  2 cos2x2dx         1+cos2A=2cos2A=cosx dx=sin x+C

(ii)
1-cos 2x2dx =2sin2x2dx            1-cos 2x=2sin2x=sinx dx=-cos x+C

Page No 19.4:

Question 3:

Evaluate: e6 loge x-e5 loge xe4 loge x-e3 loge xdx

Answer:

e6 log x-e5 logxe4 logx -e3 logxdx=elogx6-elog x5elogx4-elogx3dx=x6-x5x4-x3dx=x5x3dx=x2dx=x33+C

Page No 19.4:

Question 4:

Evaluate: 1ax bxdx

Answer:

    dxaxbx=a-x b-xdx=a-xb-x- loge ab+C

Page No 19.4:

Question 5:

Evaluate: (i) cos 2x+2 sin2 xsin2 xdx
(ii) 2 cos2 x-cos 2xcos2 xdx

Answer:

(i)
cos 2x+2 sin2xsin2xdx=1-2 sin2x+2 sin2xsin2xdx           cos 2x=1-2 sin2x=cosec2x dx=-cot x+C                                         

(ii)
2 cos2x-cos 2xcos2xdx=2 cos2x- 2 cos2x-1cos2xdx                       cos 2x=2 cos2x-1=sec2x dx=tanx+C

Page No 19.4:

Question 6:

Evaluate: elog xxdx

Answer:

 elogxxdx=xxdx=1x dx=x-12dx=x-12+1-12+1+C=2x+C



Page No 19.47:

Question 1:

11-cos 2x dx

Answer:

11-cos 2xdx= 12sin2xdx                   1-cos 2x=2sin2x= 12cosec x dx= 12ln cosec x-cotx+C= 12 ln 1sinx-cosxsinx+C= 12 ln 2sin2x2sinx+C                 1-cos x=2sin2x2= 12 ln 2sin2x22sinx2 cosx2+ C         sin x=2sinx2 cosx2= 12 ln tanx2+C

Page No 19.47:

Question 2:

11+cos x dx

Answer:

11+cosxdx= 12cos2x2dx= 12secx2 dx= 12×2 ln tanx2+secx2+C= 2 ln 1+sinx2cosx2+C= 2 ln sinx4+cosx42cos2x4-sin2x4+C                    1+sinθ=sin2θ2+cos2θ2+2sinθ2cosθ2=sinθ2+cosθ22& cosθ=cos2θ2-sin2θ2= 2 ln sinx4+cosx42cosx4-sinx4cosx4+sinx4+C= 2 ln sinx4+cosx4cosx4-sinx4+C= 2 ln 1+tanx41-tanx4+C= 2 ln tanπ4+x4+C

Page No 19.47:

Question 3:

1+cos 2x1-cos 2x dx

Answer:

1+cos2x1-cos2xdx= 2cos2x2sin2xdx= cotx dx= ln sinx+C

Page No 19.47:

Question 4:

1- cos x1+cos x dx

Answer:

1-cosx1+cosxdx= 2sin2x22cos2x2dx          1-cos x=2sin2x2  &  1+cosx=2cos2x2= tanx2 dx= -2 ln cosx2+C

Page No 19.47:

Question 5:

Evaluate the following integrals:
sec x sec 2xdx

Answer:

I=sec x sec 2xdx

=cos2xcosxdx=1-2sin2xcosxdx

=1cosxdx-2sinx·sinxcosxdx=secx·dx-2tanx·secx·dx=lnsecx+tanx-2secx+c

Page No 19.47:

Question 6:

cos 2xcos x+sin x2 dx

Answer:

Let I=cos 2xcosx+sinx2dx       = cos2x-sin2xcosx+sinx2dx       = cos x-sin xcos x+sin xdxPutting cos x+sin x=t                               -sinx+cosx=dtdxcosx-sinxdx=dt I=1tdt      = ln t+C      = ln cos x+sin x+C         t=cos x+sin x

Page No 19.47:

Question 7:

sin x-asin x-b dx

Answer:

Let I=sinx-asinx-bdxPutting x-b=t    x=b+t& dx=dt I= sinb+t-asintdt      =sinb-a+tsintdt      = sinb-acos tsint+cosb-a sin tsin tdt      = sinb-acot t dt+cosb-adt      = sinb-a ln sin t+t cosb-a+C      = sinb-a ln sinx-b+x-bcosb-a+C        t=x-b    

Page No 19.47:

Question 8:

sin x-αsin x+α dx

Answer:

Let I=sinx-αsinx+αdxPutting x+α=t    x=t-α&  dx=dtI=sin t-2αsin tdt     = sin t cos 2αsin t-cos t sin 2αsin tdt     = cos 2αdt-sin 2αcot t dt     = tcos 2α-sin 2α ln sin t+C     = x+αcos 2α-sin 2α ln sin x+α+C             t=x+α     = xcos 2α-sin 2α ln sin x+α+C

Page No 19.47:

Question 9:

1+tan x1-tan x dx

Answer:

Let I=1+tanx1-tanxdx       = 1+sinxcosx1-sinxcosxdx       = cosx+sinxcosx-sinxdxPutting cosx-sinx=t -sinx-cosxdx=dt sinx+cosxdx=-dt I= -1tdt      = - ln  t +C      = - ln cosx-sinx+C        t=cos x-sin x

Page No 19.47:

Question 10:

cos xcos x-a dx

Answer:

Let I=cosxcosx-adxPutting x-a=t             x=a+t  dx=dt I= cosa+tdtcost      = cos a cos tcos t-sin a sin tcos tdt      = cos a-sin a tan tdt      = tcos a-sin a ln sec t+C      = x-acos a-sin a ln secx-a+C        t=x-a

Page No 19.47:

Question 11:

1-sin 2x1+ sin 2x dx

Answer:

1-sin 2x1+sin 2xdx=cos2x+sin2x-2 sin x cos xcos2x+sin2x+2 sin x cos x dx= cos x-sin x2cos x+sin x2dx= cos x-sin xcos x+sin xdx= 1-tan x1+tan xdx= tan π4-xdx=1-1ln sec π4-x          tan ax+bdx=1aln  sec ax+b+C= -ln cos π4-x-1+C= ln cos π4-x+C 

Page No 19.47:

Question 12:

e3xe3x+1 dx

Answer:

Let I=e3xe3x+1dxPutting e3x+1=t    3e3x=dtdxdx=dt3e3x I= e3x3te3xdt      = 131tdt      = ln t3+C      = ln e3x+13+C

Page No 19.47:

Question 13:

sec x tan x3 sec x+5 dx

Answer:

Let I=secx tanx3 sec x+5dxPutting sec x=t        dtdx=sec x tan xdt=sec x tan x dxI= dt3t+5    = 13 ln 3t+5+C    = 13 ln 3 sec x+5+C        t=sec x

Page No 19.47:

Question 14:

1-cot x1+cot x dx

Answer:

Let I=1-cotx1+cotxdx      = 1-cosxsinx1+cosxsinxdx      = sinx-cosxsinx+cosxdxPutting  sinx+cosx=t cosx-sinxdx=dt sinx-cosxdx=-dt I= -dtt      = -ln  t +C      = -ln sinx+cosx+C

Page No 19.47:

Question 15:

sec x cosec xlog (tan x) dx

Answer:

Note: Here , we are considering log x as loge x.Let I=secx cosecxlog tanxdxPutting log tan x=t sec2xtanx=dtdxsecx cosecx dx=dt I= 1tdt      =log t+C      = log  log tanx+C

Page No 19.47:

Question 16:

1x (3+log x) dx

Answer:

Here, we are considering log x as logex .Let I=1x3+logxdxPutting log x=t 1x=dtdxdxx=dt I= dt3+t      = log 3+t+C     = log 3+log x+C       t= log x

Page No 19.47:

Question 17:

ex+1ex+x dx

Answer:

Let I=ex+1ex+xdxPutting ex+x=tex+1=dtdxex+1dx=dt I=1tdt      = ln  t +C      = ln  ex+x+C          t=ex+x

Page No 19.47:

Question 18:

1x log x dx

Answer:

Here, we are considering log x as loge x .Let I=1x log xdxPutting logx=t1x=dtdx1xdx=dt I= 1tdt      = log logx+C

Page No 19.47:

Question 19:

sin 2xa cos2 x+b sin2 x dx

Answer:

Let I=sin 2xacos2x+bsin2xdx       = sin 2xa1-sin2x+bsin2x dx       = sin 2xb-asin2x+a dxPutting sin2x=t 2sin x.cos x= dtdx sin 2x=dtdxsin 2x dx=dtI= 1b-at+adt    = 1b-a ln b-at+a+C                   1ax+bdx=1alnax+b+C    = 1b-a ln b-asin2x+a+C                  t= sin2x    = 1b-a ln bsin2x+a1-sin2x+C    = 1b-a ln bsin2x+acos2x+C

Page No 19.47:

Question 20:

cos x2+3 sin x dx

Answer:

Let I= cos x2+3sin xdxPutting  sinx=t          cosx=dtdxcosx dx=dt I= dt2+3t      = 13ln 2+3t+C                1ax+bdx=1aln ax+b+C     = 13 ln 2+3 sinx+C          t=sin x 

Page No 19.47:

Question 21:

1-sin xx+ cos x dx

Answer:

Let I=1-sinxx+cosxdxPutting  x+cosx=t1-sinx=dtdx1-sinxdx=dt I= 1tdt      =ln t+C      = ln x+cosx+C          t= x+ cos x

Page No 19.47:

Question 22:

ab+cex dx

Answer:

Let I=ab+cexdxDividing numerator and denominator by ex I=ae-xbe-x+cdxPutting e-x=t-e-x=dtdxe-xdx=-dt I=-a bt+cdt      = -ab ln bt+c+C              1ax+bdx=1aln ax+b+C      = -ab ln be-x+c+C         t=e-x+C

Page No 19.47:

Question 23:

1ex+1 dx

Answer:

Let I=1ex+1dx       =e-x1+e-xdxPutting e-x=t-e-x=dtdxe-xdx=-dt I= -11+tdt      = -ln 1+t+C      = - ln 1+e-x+C

Page No 19.47:

Question 24:

cot xlog sin x dx

Answer:

Note:Here we are considering log x as loge xLet I=cot xlog sin xdxPutting log sin x=t cot x=dtdxcot x dx=dtI= 1tdt     =log t+C            = log log sinx+C         t=log sin x    

Page No 19.47:

Question 25:

e2xe2x-2 dx

Answer:

Let I=e2xe2x-2dxPutting e2x=t2e2x=dtdxe2xdx=dt2 I= 121t-2dt      = 12 ln t-2+C      = 12 ln e2x-2+C              t= e2x

Page No 19.47:

Question 26:

2 cos x-3 sin x6 cos x+4 sin x dx

Answer:

2cosx-3sinx6cosx+4sinxdx2cosx-3sinx23cosx+2sinxdtLet, 3cosx+2sinx=t  2cosx-3sinx=dtdx  2cosx-3sinxdx=dtNow, 2cosx-3sinx23cosx+2sinxdt= dt2t=12logt+C= 12 log 3cosx+2sinx+C



Page No 19.48:

Question 27:

cos 2x+x+1x2+sin 2x+2x dx

Answer:

Let I=cos2x+x+1x2+sin2x+2xdxPutting x2+sin2x+2x=t 2x+2cos 2x+2=dtdxx+cos 2x+1dx=dt2 I= 121tdt      =12lnt+C      = 12 ln x2+sin2x+2x+C             t=x2+sin 2x+2x

Page No 19.48:

Question 28:

1cosx+a cosx+bdx

Answer:

​1cosx+a cosx+bdxMultiplying and Dividing by sinx+b-x+a, we get=1sinx+b-x+a×sinx+b-x+acosx+a cosx+bdx=1sinb-a×sinx+b-x+acosx+a cosx+bdx=1sinb-asinx+bcosx+a-sinx+acosx+bcosx+a cosx+bdx=1sinb-asinx+bcosx+bdx-sinx+acosx+adx=1sinb-atanx+bdx-tanx+adx=1sinb-alogsecx+b-logsecx+a+c=1sinb-alogsecx+bsecx+a+c

Hence, 1cosx+a cosx+bdx=1sinb-alogsecx+bsecx+a+c.

Page No 19.48:

Question 29:

-sin x+2 cos x2 sin x+cos x dx

Answer:

Let I=-sinx+2cosx2sinx+cosxdxPutting 2sinx+cosx=t2cosx-sinx=dtdx-sinx+2cosxdx=dt I=1tdt      =lnt+C      = ln 2sinx+cosx+C            t=2sin x+cos x

Page No 19.48:

Question 30:

  cos 4x-cos 2xsin 4x-sin 2x dx

Answer:

cos4x-cos2xsin4x-sin2xdx= -2sin4x+2x2sin4x-2x22cos4x+2x2sin4x-2x2dx                 cos A-cos B=-2sin A+B2sin A-B2  & sin A-sin B=2cos A+B2sin A-B2= -sin 3xcos 3xdx= -tan 3x dx= -ln sec 3x3+C= 13 ln sec 3x-1+C= 13 ln cos 3x+C

Page No 19.48:

Question 31:

sec xlog sec x+tan x dx

Answer:

Note: Here, we are considering log x as logexLet I=secxlog secx+tanxdxPutting log secx+tanx=t secx tanx+sec2xsecx+tanx=dtdx secxsecx+tanxsecx+tanx=dtdxsecx dx=dt I= dtt      =logt+C      =log log secx+tanx+C

Page No 19.48:

Question 32:

cosec xlog tanx2 dx

Answer:

Note : Here, we are considering log x as logexLet I=cosec xlog tanx2dxPutting log tan x2=t12sec2x2tanx2=dtdx12 sinx2.cosx2=dtdx1sinx=dtdxcosec x dx=dt I= dtt      =logt+C      = log log tanx2+C

Page No 19.48:

Question 33:

1x log x log log x dx

Answer:

Note: Here, we are considering log x as loge x .Let I=1x logx loglogxdxPutting  loglogx=t1xlogx=dtdx1xlogxdx=dtI=dtt     =logt+C     = logloglogx+C        t=loglogx

Page No 19.48:

Question 34:

cosec2 x1+cot x dx

Answer:

Let I=cosec2x1+cot xdxPutting cotx=t-cosec2x=dtdxcosec2x dx=-dt I= -dt1+t     =- ln 1+t+C     =- ln 1+cot x+C          t=cot x

Page No 19.48:

Question 35:

10 x9+10x loge 1010x+x10 dx

Answer:

Let I=10x9+10x loge1010x+x10dxPutting 10x+x10=t10x loge10+10x9=dtdx10x loge10+10x9dx=dt I= 1tdt      =ln t+C      = ln 10x+x10+C        t=10x+x10

Page No 19.48:

Question 36:

1-sin 2xx+cos2 x dx

Answer:

Let I=1-sin2xx+cos2xdxPutting  x+cos2x=t 1-2cosx.sinx=dtdx1-sin 2xdx=dt I=1tdt      = ln t+C      = ln x+cos2x+C        t= x+cos2x

Page No 19.48:

Question 37:

1+tan xx+log sec x dx

Answer:

Note: Here, we are considering log x as loge xLet I=1+tanxx+log secxdxPutting x+log secx=t1+secx tanxsecx=dtdx1+tanxdx=dt I= 1tdt      = log t+C       = log x+log secx+C      t=x+log sec x

Page No 19.48:

Question 38:

sin 2xa2+b2 sin2 x dx

Answer:

Let I=sin 2xa2+b2sin2xdxPutting sin2x=t2sinx.cosx=dtdxsin 2x=dtdxsin 2x dx=dt I= 1a2+b2tdt      = 1b2 ln a2+b2t+C      = 1b2 ln a2+b2 sin2x+C          t= sin2x

Page No 19.48:

Question 39:

x+1x x+log x dx

Answer:

Note: Here, we are considering log x as loge xLet I=x+1xx+logxdxPutting x+logx=t1+1x=dtdxx+1xdx=dt I= 1tdt      = log  t+C      = log x+logx+C                              

Page No 19.48:

Question 40:

11-x22+3 sin-1 x dx

Answer:

Let I=11-x22+3 sin-1xdxPutting sin-1x=t11-x2=dtdx11-x2dx=dt I= 12+3tdt      = 13 ln 2+3t+C      = 13 ln 2+3sin-1x+C        t=sin-1x

Page No 19.48:

Question 41:

sec2 xtan x+2 dx

Answer:

Let I=sec2xtanx+2dxPutting tan x=tsec2x=dtdxsec2x dx=dt I= 1t+2dt      = ln t+2+C      = ln tanx+2+C         t= tan x

Page No 19.48:

Question 42:

2 cos 2x+sec2 xsin 2x+tan x-5 dx

Answer:

Let I=2 cos 2x+sec2xsin 2x+tan x-5dxPutting  sin 2x+tan x-5=t2cos 2x+sec2x=dtdx2cos 2x+sec2xdx=dt I= 1tdt      = ln t+C      = ln sin 2x+tan x-5+C       t=sin 2x+tan x-5 

Page No 19.48:

Question 43:

sin 2xsin 5x sin 3x dx

Answer:

sin 2xsin 5x sin 3xdx= sin 5x-3xsin 5x sin 3xdx=sin 5x cos 3x-cos 5x sin 3xsin 5x sin 3xdx= sin 5x cos 3xsin 5x sin 3x-cos 5x sin 3xsin 5x sin 3xdx= cot 3x-cot 5x dx=cot 3x dx-cot 5x dx= 13 ln sin 3x-15 ln sin 5x+C  

Page No 19.48:

Question 44:

1+cot xx+log sin x dx

Answer:

Note: Here, we are considering log x as loge xLet I=1+cot xx+log sin xdxPutting x+log sinx=t1+cotx=dtdx1+cot xdx=dt I=1tdt      = log t+C      = log x+log sinx+C        t= x+log sin x

Page No 19.48:

Question 45:

1xx+1 dx

Answer:

Let I=1xx+1dxPutting x+1=t12x=dtdx1xdx=2dt I=21tdt      = 2 ln t+C      = 2 ln x+1+C           t=x+1

Page No 19.48:

Question 46:

tan 2x tan 3x tan 5x dx

Answer:

We know that, tan 5x=tan 2x+3x tan 5x=tan 2x+tan 3x1-tan 2x tan 3x tan 5x-tan 2x tan 3x tan 5x=tan 2x+tan 3xtan 2x tan 3x tan 5x = tan 5x -tan 2x-tan 3x tan 2x tan 3x tan 5x=tan 5x-tan 2x-tan 3xdx                                            = 15 ln sec 5x-12 ln sec 2x-13 ln sec 3x+C

Page No 19.48:

Question 47:

1+tan x tan x+θ dx

Answer:

Let I=1+tan x tan x+θdx       = 1+tanxtan x+tan θ1-tan x tan θdx       = 1+tan2 x1-tan x tan θdx       = sec2x dx1-tan x tan θPutting tanx=t sec2x=dtdx   dx=dtsec2x I= 11-t tanθdt      = -1tan θ ln 1-t tan θ+C           1ax+bdx=1aln ax+b+C      = -cot θ ln 1-tanx tan θ+C      = cot θ ln 11-tan x tan θ+C      = cot θ ln cosx cosθcos x cos θ-sin x sin θ+C      = cot θ ln cos xcos x+θ+C'                                       Let C'=C+cot θ ln cosθ

Page No 19.48:

Question 48:

sin 2xsin x-π6 sin x+π6 dx

Answer:

Let I=sin 2xsinx-π6 sinx+π6dx        = sin 2xsin2x-sin2π6 dx          sin A+B sinA-B=sin2A-sin2B        = sin 2xsin2x-14dxPutting sin2x-14=t 2sin x cos x dx=dt sin 2x dx=dt I= 1tdt      = ln  t +C      = ln sin2x-14+C        t=sin2x-14

Page No 19.48:

Question 49:

ex-1+xe-1ex+xe dx

Answer:

Let I=ex-1+xe-1ex+xedxPutting ex+xe=t ex+exe-1=dtdx eex-1+xe-1=dtdxex-1+xe-1dx=dte I= 1e1tdt      =1e ln t+C      = 1e ln ex+xe+C            t=ex+xe 

Page No 19.48:

Question 50:

1sin x cos2 x dx

Answer:

1sin x cos2xdx=sin2x+cos2xsin x cos2xdx=tan x sec x+cosec x dx= sec x+ln cosec x-cot x+C= sec x+ln tanx2+C                                    cosecx-cotx=1-cosxsin x=tanx2

Page No 19.48:

Question 51:

1cos 3x-cos x dx

Answer:

1cos 3x-cos xdx= 14cos3x-4cosxdx                 cos 3x=4 cos 3x-3 cos x= 14cos xcos2x-1dx= -141cos x sin2xdx= -14sin2x+cos2xcos x sin2x dx= -14sec x dx+cot x cosec x dx= -14ln secx+tanx-cosec x+C= 14cosec x-lnsec x+tan x+C



Page No 19.57:

Question 1:

log xx dx

Answer:

log xxdxLet, log x=t1x=dtdxNow, log xxdx=t·dt=t22+C=log x22+C 

Page No 19.57:

Question 2:

log1+1xx 1+x dx

Answer:

log 1+1xx1+xdxLet, log 1+1x=t11+1x×-1x2=dtdxxx+1×-1x2=dtdx-dxxx+1=dtdxxx+1=-dtNow, log 1+1xx1+xdx=t ·-dt=-t22+C=-12log1+1x2+C

Page No 19.57:

Question 3:

1+x2x dx

Answer:

1+x2xdxLet, 1+x=t12x=dtdxdxx=2dtNow, 1+x2xdx =2t2dt=23t3+C=231+x3+C

Page No 19.57:

Question 4:

1+ex ex dx

Answer:

1+ex·exdxLet 1+ex=tex=dtdxexdx=dtNow, 1+ex·exdx=t·dt=t12+112+1+C=23t32+C=231+ex32+C

Page No 19.57:

Question 5:

cos2 x3sin x dx

Answer:

cos2 x13 sin x dxLet, cos x=t-sin x=dtdxsinx dx=-dtNow, cos2 x13 sin x dx=-t23dt=-t23+123+1+C=-35 t53+C=-35cos53x+C

Page No 19.57:

Question 6:

ex1+ex2 dx

Answer:

ex dx1+ex2Let 1+ex=tex =dtdxex dx=dtNow, ex dx1+ex2=dtt2=t-2 dt=t-2+1-2+1+C=-1t+C=-11+ex+C

Page No 19.57:

Question 7:

cot3 x cosec2 x dx

Answer:

cot3x cosec2x dxLet, cot x=t-cosec2x =dtdxcosec2x dx=-dtNow, cot3x cosec2x dx=t3-dt=-t44+C=-cot4x4+C

Page No 19.57:

Question 8:

esin-1 x21-x2 dx

Answer:

esin-1 x21-x2 dxLet esin-1 x=tDifferentiating both sides w.r.t. x,esin-1 x ×11-x2 dx=dtNow, esin-1 x21-x2 dx=esin-1 x·esin-1 x1-x2dx =t ·dt=t22+C=esin-1 x22+C

Page No 19.57:

Question 9:

1+sin xx-cos x dx

Answer:

1+sin xx-cos xdxLet, x-cos x=t1+sin x=dtdx1+sin x dx=dtNow, 1+sin xx-cos xdx=dtt=t-12dt=t-12+1-12+1+C=2t+C=2x-cos x+C

Page No 19.57:

Question 10:

11-x2sin-1 x2 dx

Answer:

dx1-x2sin-1 x2Let, sin-1 x=t11-x2=dtdx11-x2 dx=dtNow,dx1-x2sin-1 x2=dtt2=t-2dt=t-2+1-2+1+C=-1t+C=-1sin-1 x+C



Page No 19.58:

Question 11:

cot xsin x dx

Answer:

     cot xsin xdx=cos xsin x sin x dx=cos xsin x32dxLet sin x=tcos x=dtdxcos x dx=dtNow, cos xsin x32dx=dtt32=t-32dt=t-32+1-32+1+C=-2t+C=-2sin x+C

Page No 19.58:

Question 12:

tan xcos x dx

Answer:

     tan xcos xdxsin xcos x cos x dxsin xcos 32xdxLet cos x=t-sin x dx=dtsin x =-dtdxNow, sin xcos 32xdx=-1t32dt=-t-32dt=-t-32+1-32+1+C=2t+C=2cos x+C

Page No 19.58:

Question 13:

cos3 xsin x dx

Answer:

     cos3 xsin xdx=cos2x·cos xsin x dx=1-sin2x cos xsin xdxLet sin x=tcos x=dtdxcos x dx=dtNow, 1-sin2x cos xsin xdx=1-t2t·dt=1t-t32dt=t-12-t32dt=t-12+1-12+1-t32+132+1+C=2t-25t52+C=2sinx-25sin52 x+C

Page No 19.58:

Question 14:

sin3 xcos x dx

Answer:

     sin3 xcos xdx=sin2x·sin xcos x dx=1-cos2x sin xcos xdxLet cos x=t-sin x=dtdxsin x dx=-dtNow,1-cos2xsin xcos xdx=-1-t2tdt=t2-1tdt=t32-t-12dt=t32+132+1-t-12+1-12+1+C=25t52-2t+C=25 cos52x -2 cos x+C

Page No 19.58:

Question 15:

1tan-1 x.1+x2 dx

Answer:

dxtan-1x 1+x2Let tan-1x=t11+x2=dtdx11+x2dx=dt

Now, dxtan-1x 1+x2= dtt=t-12dt=t-12+1-12+1+C=2 t+C=2 tan-1 x+C

Page No 19.58:

Question 16:

tan xsin x cos x dx

Answer:

tan xsin x cos xdx=tan xsin xcos x×cos2xdx=tan xtan x×sec2 x dx=1tan x×sec2 x dx=tan x-12sec2 x dxLet tan x=tsec2x=dtdxsec2x dx=dtNow, tan x-12sec2 x dx= t-12dt=t-12+1-12+1+C=2 t+C=2 tan x+C

Page No 19.58:

Question 17:

1xlog x2 dx

Answer:

1x log x2 dxLet log x=t1x dx=dtNow, 1x log x2 dx=t2dt=t33+C=log x33+C

Page No 19.58:

Question 18:

sin5 x cos x dx

Answer:

sin5 x cos x dxLet sin x=tcos x=dtdxcos x dx=dtNow, sin5 x cos x dx=t5dt=t66+C=16 sin6 x+C

Page No 19.58:

Question 19:

tan3/2 x sec2 x dx

Answer:

tan32x·sec2x dxLet tan x=tsec2x=dtdxsec2x dx=dtNow, tan32x·sec2x dx = t32 dt=t32+132+1+C=25 t52+C=25 tan52 x+C

Page No 19.58:

Question 20:

x3x2+13 dx

Answer:

     x3x2+13 dx=x2.xx2+13dxLet x2+1=tx2=t-12x dx=dtx dx=dt2Now, x2.xx2+13dx=12t-1t3dt=121t2-1t3 dt=12t-2-t-3dt=12t-2+1-2+1-t-3+1-3+1+C=12-1t+12t2+C=12-1x2+1+12x2+12+C=12-2 x2+1+12 x2+12=14-2x2-2+1x2+12=-141+2x2x2+12+C

Page No 19.58:

Question 21:

4x+2x2+x+1 dx

Answer:

     4x+2 x2+x+1 dx=22x+1 x2+x+1 dxLet x2+x+1=t2x+1=dtdx2x+1 dx=dtNow, 22x+1 x2+x+1 dx=2t dt=2t12dt=2 t12+112+1+C=2×23t32+C=43 t32+C=43x2+x+132+C

Page No 19.58:

Question 22:

4x+32x2+3x+1 dx

Answer:

4x+32x2+3x+1dxLet 2x2+3x+1=t4x+3=dtdx4x+3 dx=dtNow, 4x+32x2+3x+1dx=dtt=t-12dt=t-12+1-12+1+C=2 t+C=2 2x2+3x+1+C

Page No 19.58:

Question 23:

11+x dx

Answer:

       dx1+x=x dxx 1+xLet 1+x=tx=t-112x=dtdxdxx =2dt

Now, xx1+xdx=2t-1tdt=21-1tdt=2 t-log t+C=2 1+x-2 log 1+x+CLet C + 2 =C'=2x-2 log 1+x+C'

Page No 19.58:

Question 24:

ecos2x sin2x dx

Answer:

Let I=ecos2x sin2x dx              Let cos2x=t             On differentiating both sides,we get              -2 cosx sinx dx=dt I=et 2 sinx cosx dt-2 sinx cosx      =-et dt      =-et +c      =-ecos2x +c

Page No 19.58:

Question 25:

1+cos xx+sin x3 dx

Answer:

1+cos xx+sin x3dxLet x+sin x=t1+cos x=dtdx1+cos x dx=dtNow, 1+cos xx+sin x3dx=dtt3=t-3dt=t-3+1-3+1+C=-12 t2+C=-12 x+sin x2+C

Page No 19.58:

Question 26:

cos x-sin x1+sin 2x dx

Answer:

      cos x-sin x1+sin 2xdxcos x-sin xcos2 x+sin2 x+2 sin x.cos x dxcos x-sin xcos x+sin x2dxLet cos x+sin x=t-sin x+cos x=dtdx-sin x+cos x dx=dtNow, cos x-sin xcos x+sin x2dx=dtt2=t-2dt=t-2+1-2+1+C=-1t+C=-1sin x +cos x+C

Page No 19.58:

Question 27:

sin 2xa+b cos 2x2 dx

Answer:

      sin 2xa+b cos 2x2dxLet a+b cos2x=t-b sin 2x dx × 2=dtsin 2x dx=-dt2bNow,  sin 2xa+b cos 2x2dx=-12bdtt2=-12bt-2dt=-12bt-2+1-2+1+C=12b×1t+C=12b a+b cos 2x+C

Page No 19.58:

Question 28:

log x2x dx

Answer:

      log x2 dxx=2 log xx dx=2log xxdxLet log x=t1x=dtdx1x dx=dtNow, 2log xxdx=2t dt=2t22+C=t2+C=log x2+C

Page No 19.58:

Question 29:

sin x1+cos x2 dx

Answer:

      sin x1+cos x2dxLet 1+cos x=t-sin x=dtdxsin x dx=-dtNow, sin x1+cos x2dx=-dtt2=-t-2 dt=-t-2+1-2+1+C=1t+C=11+cos x+C

Page No 19.58:

Question 30:

cot x·log sin x dx

Answer:

cot x·log sin x dxLet log sin x=t1sin x×cos x=dtdxcot x dx=dtNow, cot x·log sin x dx=t·dt=t22+C=log sin x22+C

Page No 19.58:

Question 31:

sec x·log sec x+tan x dx

Answer:

sec x·log sec x+tan x dxLet log sec x+tan x=tsec x tan x+ sec2xsec x+tan x=dtdxsec x sec x+tan xsec x+tan x dx=dtNow, sec x·log sec x+tan x dx=t·dt=t22+C=log sec x+ tan x22+C

Page No 19.58:

Question 32:

cosec x log cosec x-cot x dx

Answer:

cosec x·log cosec x-cot x dxLet log cosec x-cot x=t-cosec x cot x+ cosec2xcosec x-cot x=dtdxcosec x-cot xcosec x-cot x×cosec x dx=dtcosec x dx=dtNow, cosec x·log cosec x-cot x dx=t·dt=t22+C=log cosec x- cot x22+C

Page No 19.58:

Question 33:

x3 cos x4 dx

Answer:

x3·cos x4 dxLet x4=t4x3 dx=dtx3 dx=dt4Now, x3·cos x4 dx=14cos t dt=14sin t+C=14sinx4+C

Page No 19.58:

Question 34:

x3 sin x4 dx

Answer:

x3·sin x4 dxLet x4=t4x3=dtdxx3 dx=dt4Now, x3·sin x4 dx=14sin t dt=14-cos t+C=14-cos x4+C

Page No 19.58:

Question 35:

x sin-1 x21-x4 dx

Answer:

 x sin-1 x21-x4dxLet sin-1 x2=t1×2x1-x4=dtdxx dx1-x4=dt2Now,  x sin-1 x21-x4dx=12tdt=t24+C=sin-1 x224+C

Page No 19.58:

Question 36:

x3 sin x4+1 dx

Answer:

x3·sin x4+1 dxLet x4+1=t4x3=dtdxx3 dx=dt4Now, x3·sin x4+1 dx=14sin t dt=14-cos t+C=14-cos x4+1+C

Page No 19.58:

Question 37:

x+1 excos2 xex dx

Answer:

x+1 excos2 x·ex dxLet x ex=t1·ex+x ex=dtdxx+1 ex dx=dtNow, x+1 excos2 x·ex dx=dtcos2t=sec2 t dt=tan t+C=tan x ex+C

Page No 19.58:

Question 38:

x2 ex3 cos ex3 dx

Answer:

x2·ex3·cos ex3 dxLet ex3=tex3·3x2 dx=dtex3·x2 dx=dt3Now, x2·ex3·cos ex3 dx=13cost dt=13sin t +C=13sin ex3+C

Page No 19.58:

Question 39:

2x sec3 x2+3 tan x2+3 dx

Answer:

     2x sec3 x2+3·tan x2+3 dx=sec2 x2+3·sec x2+3·tan x2+3·2x dxLet sec x2+3=tsec x2+3·tan x2+3·2x=dtdxsec x2+3·tan x2+3·2x dx=dtNow, sec2 x2+3·sec x2+3·tan x2+3·2x dx=t2 dt=t33+C=sec3 x2+33+C

Page No 19.58:

Question 40:

x+1xx+log x2 dx

Answer:

x+1x·x+log x2 dxLet x+log x=t1+1x=dtdxx+1x dx=dtNow, x+1x·x+log x2 dx=t2 dt=t33+C=x+log x33+C

Page No 19.58:

Question 41:

tan x sec2 x1-tan2 x dx

Answer:

tan x·sec2 x 1-tan2x dxLet tan x=tsec2 x dx=dtNow, tan x·sec2 x 1-tan2x dx=t·1-t2dtAgain let t2=p2t dt=dpt dt=dp2Again, t·1-t2dt=121-p dp=121-p12dp=121-p12+112+1 -1+C=12×-231-p32+C=-131-p32+C=-131-tan2x32+C

Page No 19.58:

Question 42:

log xsin 1+log x2x dx

Answer:

log x sin 1+log x2x dxLet 1+log x2=t2 log x ×1x dx=dtlog xxdx=dt2Now, log x sin 1+log x2x dx=12sin t dt=12-cos t+C=-12cos 1+log x2+C

Page No 19.58:

Question 43:

1x2cos21x dx

Answer:

1x2·cos2 1x dxLet 1x=t-1x2=dtdx1x2dx=-dtNow, 1x2·cos2 1x dx=-cos2 t dt=-1+cos 2t2dt=-121+cos 2tdt=-12t+sin 2t2+C=-121x+sin 2x2+C=-121x-14sin 2x+C

Page No 19.58:

Question 44:

sec4 x tan x dx

Answer:

 sec4 x.tan x dx= sec2 x.sec2 x tan x dx= 1+tan2 x sec2 x. tan x dx= tan x+tan3 x sec2 x dxLet tan x=tsec2 x dx=dtNow,  tan x+tan3 x sec2 x dx=t+t3 dt=t22+t44+C=tan2 x2+tan4 x4+C

Page No 19.58:

Question 45:

ex cos exx dx

Answer:

ex·cos exxdxLet ex=tex×12x=dtdxexxdx=2dtNow, ex·cos exxdx=2cos t dt =2 sin t +C=2 sin ex+C

Page No 19.58:

Question 46:

cos5 xsin x dx

Answer:

cos5 xsin xdx= cos4 x .cos xsin xdx=cos2 x2.cos xsin xdx=1-sin2 x2×cos xsin xdx= 1-sin4 x-2 sin2 xsin xcos x dxLet sin x=tcos x dx=dtNow,  1-sin4 x-2 sin2 xsin xcos x dx= 1+t4-2t2tdt=1t+t3-2tdt=log t +t44-2t22+C=log t+t44-t2+C=log sin x+sin4 x4-sin2 x+C



Page No 19.59:

Question 47:

sinxx dx

Answer:

sin xxdxLet x=t12x=dtdxdxx=2dtNow, sin xxdx=2sin t dt=2 -cos t+C=-2 cos x+C

Page No 19.59:

Question 48:

x+1 exsin2 x ex dx

Answer:

x+1 exsin2 x exdxLet x ex=t1·ex+x ex=dtdxx+1 ex dx=dtNow, x+1 exsin2 x exdx=dtsin2 t=cosec2 t dt =-cot t+C=-cot x ex+C

Page No 19.59:

Question 49:

5x+tan-1 x.x2+2x2+1 dx

Answer:

5x+tan-1x·x2+2x2+1dxLet x+tan-1 x=t1+11+x2=dtdxx2+1+1x2+1dx=dtx2+2x2+1dx=dtNow, 5x+tan-1x·x2+2x2+1dx=5t dt=5tlog 5+C=5x+ tan1xlog 5+C

Page No 19.59:

Question 50:

em sin-1 x1-x2 dx

Answer:

em sin-1x1-x2dxLet sin-1 x=t11-x2dx=dtNow, em sin-1x1-x2dx=em t·dt=emtm+C=em sin-1xm+C

Page No 19.59:

Question 51:

cosxx dx

Answer:

cos xxdxLet x=t12x=dtdxdxx=2 dtNow, cos xxdx=2cos t dt=2 sin t+C=2 sin x+C

Page No 19.59:

Question 52:

sin tan-1 x1+x2 dx

Answer:

sin tan-1 x1+x2dxLet tan-1 x=t11+x2dx=dtNow, sin tan-1 x1+x2dx=sin t dt=- cos t+C=- cos tan-1 x+C

Page No 19.59:

Question 53:

sin log xx dx

Answer:

sin log xxdxLet log x=t1xdx=dtNow, sin log xxdx=sin t dt=- cos t+C=- cos log x+C

Page No 19.59:

Question 54:

em tan-1 x1+x2 dx

Answer:

em tan-1 x1+x2dxLet tan-1 x=t11+x2dx=dtNow, em tan-1 x1+x2dx=emt dt=emtm+C=em tan-1xm+C

Page No 19.59:

Question 55:

xx2+a2+x2-a2 dx

Answer:

x dxx2+a2+x2-a2Let x2=t2x=dtdxx dx=dt2Now, x dxx2+a2+x2-a2=12dtt+a2+t-a2=12dtt+a2+t-a2×t+a2-t-a2t+a2-t-a2=12t+a2-t-a2t+a2-t-a2dt=14a2t+a212dt-14a2t-a212dt=14a2t+a212+112+1-14a2t-a212+112+1+C=16a2t+a232-t-a232+C=16a2x2+a232-x2-a232+C

Page No 19.59:

Question 56:

xtan-1 x21+x4 dx

Answer:

x tan-1 x21+x4 dxLet tan-1 x2=t11+x22×2x=dtdxx dx1+x4=dt2Now, x tan-1 x21+x4 dx=12t. dt=12×t22+C=tan-1 x224+C

Page No 19.59:

Question 57:

sin-1 x31-x2 dx

Answer:

sin-1 x31-x2 dxLet sin-1 x=t11-x2dx=dtNow, sin-1 x31-x2 dx = t3 dt=t44+C=sin-1 x44+C

Page No 19.59:

Question 58:

sin 2+3 log xx dx

Answer:

 sin 2+3 log xxdxLet 2+3 log x=t3x=dtdxdxx=dt3Now,  sin 2+3 log xxdx=13 sin t dt=13 -cos t+C=-13cos 2+3 log x+C

Page No 19.59:

Question 59:

x ex2 dx

Answer:

 x.ex2 dxLet x2=t2x dx=dtx dx=dt2Now,  x.ex2 dx=12et dt=12et+C=12ex2+C

Page No 19.59:

Question 60:

e2x1+ex dx

Answer:

e2x dx1+exex.ex1+exdxLet 1+ex=t ex=t-1ex dx=dtNow, ex.ex1+exdx= t-1.dtt=1-1tdt=t-log t+C=1+ex-log 1+ex+CLet C+1=C'=ex-log 1+ex+C'

Page No 19.59:

Question 61:

sec2xx dx

Answer:

 sec2 xxdxLet x=t12x=dtdxdxx=2dtNow,  sec2 xxdx=2 sec2 t dt=2 tan t+C=2 tan x+C

Page No 19.59:

Question 62:

tan3 2x sec 2x dx

Answer:

 tan3 x 2x.sec 2xdx= tan2 2x. sec 2x tan 2x dx= sec2 2x-1 sec 2x tan 2x dxLet sec 2x=tsec 2x tan 2x×2=dtdxsec 2x tan 2x dx=dt2Now,  tan3 x 2x.sec 2xdx=12t2-1 dt=12t33-t+C=12 sec3 2x3-sec2x+C=16 sec3 2x-sec 2x2+C

Page No 19.59:

Question 63:

x+x+1x+2 dx

Answer:

We have,I=x+x+1x+2 dxLet, x+1=t2Differentiating both sides we getdx=2tdtNow, integration becomesI=t2-1+tt2+12t dt=2t3+t2-tt2+1 dt=2t3+t-t+t2+1-1-tt2+1 dt=2t3+t+t2+1-t-t-1t2+1 dt=2t3+tt2+1 dt++2t2+1t2+1 dt+2-2t-1t2+1 dt=2tdt+2dt-22tt2+1 dt-21t2+1 dt=t2+2t-2log t2+1-2tan-1t+C=x+1+2x+1-2log x+2-2tan-1x+1+C

Page No 19.59:

Question 64:

555x 55x 5x dx

Answer:

555x·55x·5x dxLet 5x=t5x log 5=dtdx5x dx=dtlog 5Now, 555x·55x·5x dx=55t·5t· dtlog 5Again let 5t=p5t log 5=dpdt5t dt=dplog 5Again 55t·5t· dtlog 5=5p·dplog 52=5plog 53+C=555xlog 53+C

Page No 19.59:

Question 65:

1xx4-1 dx

Answer:

dxxx4-1=x dxx2x22-1Let x2=t2x=dtdxx dx=dt2Now, x dxx2x22-1=12dttt2-1=12 sec-1 t+C=12sec-1 x2+C

Page No 19.59:

Question 66:

ex-1 dx

Answer:

ex-1dxLet ex-1=t2ex=t2+1ex=2t dtdxdx=2t dtexdx=2t dtt2+1Now, ex-1dx=t·2t dtt2+1=2t2 dtt2+1=2t2+1-1t2+1dt =2dt-2dtt2+1=2t -2 tan-1 t+C=2ex-1-2 tan-1 ex-1+C

Page No 19.59:

Question 67:

1x+1x2+2x+2 dx

Answer:

Let I=dxx+1 x2+2x+2=dxx+1 x2+2x+1+1=dxx+1 x+12+1Putting x+1=tdx=dtNow, integral becomesI=dtt t2+1=t·dtt2 t2+1Again putting t2=p2t dt=dpt dt=dp2Now, integral becomesI=12 dpp p+1=12dpp2+p=12dpp2+p+14-14=12dpp+122-122=12 12×12 log p+12-12p+12+12 +C=12 log  pp+1+C=12 log t2t2+1+C=12 log x+12x+12+1+C= log x+12x+12+1+C= log x+1x2+2x+2+C

Page No 19.59:

Question 68:

x51+x3 dx

Answer:

x5 dx1+x3=x3.x2 dx1+x3Let 1+x3=t x3=t-13x2=dtdxx2 dx=dt3Now, x3.x2 dx1+x3=13t-1t dt=13t-1tdt=13 t12-t-12dt=13t12+112+1-t-12+1-12+1+C=1323t32-2t+C=29 1+x332-231+x312+C

Page No 19.59:

Question 69:

4x35-x2 dx

Answer:

 4x3 5-x2 dx=4x2×x 5-x2 dxLet 5-x2=t x2=5-t2x=-dtdxx dx=-dt2Now, 4x2×x 5-x2 dx=4-2 5-t.t dt=-25t12+2 t32 dt=-10 t12+112+1+2 t32+132+1+C=-203t32+45t52+C=-2035-x232+455-x252+C

Page No 19.59:

Question 70:

1x+x dx

Answer:

dxx+x=dxx 1+xLet 1+x=t12x=dtdxdxx=2dtNow, dxx 1+x=2dtt=2dtt=2 log t+C=2 log 1+x+C

Page No 19.59:

Question 71:

1x2 x4+13/4 dx

Answer:

dxx2x4+134=dxx2 x41+1x434= dxx2.x3 1+1x434=1+1x4-34x5dxLet 1+1x4=t-4x5dx=dtdxx5=-dt4Now, 1+1x4-34x5dx=-14 t-34 dt=-14 t-34+1-34+1+C=-t14+C=-1+1x414+C

Page No 19.59:

Question 72:

sin5 xcos4 x dx

Answer:

sin5 xcos4 xdx=sin4 x. sin xcos4 xdx=sin2 x2. sin xcos4 xdx= 1-cos2 x2 sin xcos4 xdx= 1+cos4 x-2 cos2 xcos4 xsin x dx= 1cos4 x+1-2cos2 xsin x dxLet cos x=t-sin x=dtdxsin x dx=-dtNow,  1cos4 x+1-2cos2 xsin x dx=- t-4+1-2t-2dt=-t-4+1-4+1+t-2t-2+1-2+1+C=--13t3+t+2t+C=13t3-t-2t+C=13 cos3 x-cos x-2cos x+C



Page No 19.65:

Question 1:

x2 x+2 dx

Answer:

x2 x+2 dxLet x+2=tx=t-2dx=dtNow, x2 x+2 dx=t-22 t dt=42-4t+4t12 dt=t2+12-4t1+12+4t12dt=t52-4t32+4t12dt=t52+152+1-4t32+132+1+4t12+112+1+C=27t72-85t52+83t32+C=27x+272-85x+252+83x+232+C

Page No 19.65:

Question 2:

x2x-1 dx

Answer:

x2x-1dxLet x-1=t2x=t2+11=2t dtdxdx=2t dtNow, x2x-1dx=t2+12t2t dt=2t4+2t2+1dt=2t4+14+1+2t2+12+1+t+C=2t55+2t33+t+C=23t5+10t3+15t15+C=215t3t4+10t2+15+C=215x-1 3x-12+10x-1+15+C=215x-1 3x2-2x+1+10x-10+15+C=215x-1 3x2-6x+3+10x-10+15+C=215x-13x2+4x+8 +C

Page No 19.65:

Question 3:

x23x+4 dx

Answer:

x2 dx3x+4Let 3x+4=t x=t-431=13.dtdxdx=dt3Now, x2 dx3x+4=13t-432tdt=127t2t-8tt+16tdt=127t32-8t12+16t-12dt=127 t32+132+1+8t12+112+1+16t-12+1-12+1+C=127 25t52-8×23t32+32t12+C=2135t52-1681t32+3227t12+C=21353x+452-16813x+432+32273x+412+C

Page No 19.65:

Question 4:

2x-1x-12 dx

Answer:

2x-1x-12dxLet x-1=tx=t+11=dtdxNow, 2x-1x-12dx=2t+1-tt2dt=2t+1t2dt=2dtt+t-2 dt=2 log t+t-2+1-2+1+C=2 log x-1-1x-1+C

Page No 19.65:

Question 5:

2x2+3 x+2 dx

Answer:

2x2+3 x+2 dxLet x+2=tx=t-2dx=dt2t-22+3t dt=2t t2-4t+4+3tdt=2t52-4t32+4t12 dt+3t12 dt=2t52+152+1-4t32+132+1+4t12+112+1+3t12+112+1+C=227t72-85t52+83t32+2t32+C=47t72-165t52+163t32+2t32+C=47t72-165t52+223t32+C=47x+272-165x+252+223x+232+C

Page No 19.65:

Question 6:

x2+3x+1x+12 dx

Answer:

x2+3x+1x+12 dxLet x+1=tx=t-11=dtdxdx=dtNow, x2+3x+1x+12 dx=t-12+3t-1+1t2dt=t2-2t+1+3t-3+1t2dt=t2+t-1t2dt=1+1t-t-2 dt=t+log t-t-2+1-2+1+C=t+log t+1t+C=x+1+log x+1+1x+1+CLet 1+C=C'=x+log x+1+1x+1+C'

Page No 19.65:

Question 7:

x21-x dx

Answer:

x21-xdxLet 1-x=t x=1-t1=-dtdxdx=-dtNow, x21-xdx=1-t2tdt=1-t2-2ttdt=1t+t2t-2ttdt=t-12+t32-2t12dt=t-12+1-12+1+t32+132+1-2t12+112+1+C=2t12+25t52-43t32+C=2t121+t25-23t+C=2t1215+3t2-10t15+C=21-x 15+31-x2-101-x15+C=2151-x 15+312+x2-2x-10+10x+C=2151-x 15+3+3x2-6x-10+10x+C=2151-x 3x2+4x+8+C

Page No 19.65:

Question 8:

x 1-x23 dx

Answer:

x 1-x23 dxLet 1-x=t x=1-t1=-dtdxdx=-dtNow, x 1-x23 dx=-1-t·t23 dt=-t23-t24dt=t24-t23 dt=t2525-t2424+C=24t25-25t24600+C=t2460024t-25+C=1-x24600 241-x-25+C=-1600 1-x24 1+24x+C

Page No 19.65:

Question 9:

1x+x4dx

Answer:

Let I=1x+x4dx              Let x=t4             On differentiating both sides,we get              dx=4t3 dt I=4t3t4+t44dt      =4t3t2+tdt      =4t2t+1dt      =4t-1t+1+1t+1dt      =4t-1+1t+1dt      =4t22-t+logt+1+c      =2x-4x4+4 logx4+1+cHence, 1x+x4dx=2x-4x4+4 logx4+1+c

Page No 19.65:

Question 10:

1x13 x13-1dx

Answer:

Let I=1x13 x13-1dx       =1x23- x13dx              Let x=t3             On differentiating both sides,we get              dx=3t2 dt I=3t2t323-t313dt      =3t2t2-tdt      =3tt-1dt      =3t-1+1t-1dt      =31+1t-1dt      =3t+logt-1+c      =3x13+3 logx13-1+cHence, 1x13 x13-1dx=3x13+3 logx13-1+c



Page No 19.69:

Question 1:

tan3 x sec2 x dx

Answer:

∫ tan3 x sec2 x dx
Let tan x = t
⇒ sec2 x dx = dt
Now, ∫ tan3 x sec2 x dx
= â€‹∫ t3.dt
=t44+C=tan4 x4+C

Page No 19.69:

Question 2:

tan x sec4 x dx

Answer:

∫ tan x. sec4 x dx
= ​∫ tan x. sec2 x . sec2 x dx
= ∫ tan x (1 + tan2 x) sec2 x dx
Let tan x = t
⇒ sec2 x dx = dt
Now, ∫ tan x (1 + tan2 x) sec2 x dx
= ∫ t (1 + t2) dt
= ∫ (t + t3) dt
=t22+t44+C=12tan2 x+14 tan4 x+C

Page No 19.69:

Question 3:

tan5 x sec4 x dx

Answer:

∫ tan5 x sec4 x dx
= ∫ tan5 x. sec2 x . sec2 x dx
= ∫ tan5 x (1 + tan2 x) sec2 x dx
Let tan x = t
⇒ sec2 x dx = dt
Now, ∫tan5x (1+tan2 x) sec2 x dx
= ∫ t5 (1 + t2) dt
= ∫ (t5 + t7) dt
=t66+t88+C=tan6 x6+tan8 x8+C

Page No 19.69:

Question 4:

sec6 x tan x dx

Answer:

∫ sec6 x  tan x dx
=∫ sec6 x.sec x tan x dx
​Let sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec6 x.sec x tan x dx
= ∫ t6. dt
​=t66+C=sec6 x6+C

Page No 19.69:

Question 5:

tan5 x dx

Answer:

∫ tan5 x dx
= ∫ tan4 x. tan x dx
= ∫(sec2 x – 1)2 . tan x dx
= ​​∫ (sec4 x – 2 sec2 x + 1) tan x dx
= ∫ tan x . sec4 x dx – 2 â€‹∫ sec2 x . tan x dx+  â€‹∫ tan x dx
= ∫ sec2 x. sec2 x . tan x dx – 2 â€‹∫ tan x sec2 x dx + â€‹∫ tan x dx
= ∫ (1 + tan2 x) . tan x . sec2 x dx – 2 â€‹∫ tan x . sec2 x dx + â€‹∫ tan x dx
Let I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 â€‹∫ tan x . sec2 x dx
And I2=∫ tan x dx
∫ tan5 x dx=I1 + I2
Now, I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 â€‹∫ tan x . sec2 x dx
Let tan x = t
⇒ sec2x dx = dt
I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 â€‹∫ tan x . sec2 x dx
∫ (1 + t2) . t. dt – 2 â€‹∫ t. dt
∫ (t + t3) dt – 2 â€‹∫ t dt
t22+t44-2t22+C1=t44-t22+C1=tan4 x4-tan2 x2+C1
And I2=∫ tan x dx
          =logsec x+C2
tan5 x dx=tan4 x4-tan2x2+C1+logsec x+C2                 =tan4 x4-tan2x2+logsec x+C1+C2                 =tan4 x4-tan2x2+logsec x+C      C= C1+C2

Page No 19.69:

Question 6:

tan x sec4 x dx

Answer:

 tan x ·sec4 x dx=tan x · sec2 x ·sec2 x dx=tan x·1+tan2 x sec2 x dxLet tan x=tsec2 x dx=dtNow, tan x·1+tan2 x sec2 x dx=t 1+t2 dt=t+t52dt=t12+t52dt=23t32+27t72+C=23tan32 x+27tan72 x+C

Page No 19.69:

Question 7:

sec4 2x dx

Answer:

∫ sec4 2x dx
=​ ∫ sec2 2x . sec2 2x dx
= ∫ (1 + tan2 2x) . sec2 2x dx
Let tan 2x = t
⇒sec2 2x . 2 dx = dt
sec2 2x . dx=dt2Now, 1+tan22x.sec22x dx=121+t2 dt=12t+t33+C=t2+t36+C=tan 2x2+tan3 2x6+C

Page No 19.69:

Question 8:

cosec4 3x dx

Answer:

​∫ cosec4 3x dx
= ∫ cosec2 3x . cosec2 3x dx
= ∫ (1 + cot2 3x) cosec2 3x dx
Let cot (3x) = t
⇒–cosec2 (3x) × 3 dx = dt
cosec2 3xdx=-dt3Now, 1+cot2 3x=-131+t2 dt=-13 t+t33+C=-t3-t39+C=-cot 3x3-cot3 3x9+C 

Page No 19.69:

Question 9:

cotn cosec2 x dx, n-1

Answer:

∫ cotn x cosec2 x dx
Let cot x = t
⇒ –cosec2 x dx = dt
⇒ cosec2 x dx = –dt
Now, cotnx cosec2x dx=-tn dt =-tn+1n+1+C=-cotn+1 xn+1+C

Page No 19.69:

Question 10:

cot5 x cosec4 x dx

Answer:

∫ cot5 x . cosec4 x dx
= ∫ cot5 x . cosec2 x . cosec2 x dx
= ​​∫ cot5 x . (1 + cot2 x) . ​cosec2 x dx
Let cot x = t
⇒ – cosec2 x dx = dt
⇒ cosec2 x dx = –dt
Now, ∫ cot5 x . cosec4 x dx
= ∫ t5 (1 + t2) dt
= ∫(t5 + t7) dt
=-t66+t88+C=-cot6 x6+cot8 x8+C

Page No 19.69:

Question 11:

cot5 x dx

Answer:

∫ cot5 x dx
= ∫ cot4 x . cot x dx
= ∫ (cosec2 x – 1)2 cot x dx
= ∫ (cosec4 x – 2 cosec2 x + 1) cot x dx
= ∫ cosec4 x . cot x dx – 2 â€‹∫ cot x . cosec2 x dx + â€‹∫ cot x dx
= ∫ cosec2 x . cosec2 x . cot x . dx – 2 â€‹∫ cot x cosec2 x dx + ∫​ cot x dx
=∫ (1 + cot 2 x) . cot x . cosec2 x dx – 2 â€‹∫ cot x cosec2 x dx + â€‹∫ cot x dx
= ∫ (cot x + cot3 x) cosec2 x dx – 2 â€‹∫ cot x cosec2 x dx + â€‹∫ cot x dx
Now, let I1= ∫ (cot x + cot3 x) cosec2 x dx – 2 â€‹∫ cot x cosec2 x dx
And I2= ∫ cot x dx
First we integrate I1
I1= ∫ (cot x + cot3 x) cosec2 x dx – 2 â€‹∫ cot x cosec2 x dx
Let cot x = t
⇒ – cosec2 x dx = dt
⇒ cosec2 x dx = – dt

I1= ∫ (t + t3) (– dt) – 2​∫ t (–dt)
= –∫(t + t3) + 2​∫t dt
=-t22-t44+2.t22+ C1=t22-t44+C1=cot2 x2-cot4 x4+C1
Now we integrate I2
I2= ∫ cot x dx
   = logsin x+C2

Now, ∫ cot5 x dx=I1 + I2
                        = -14cot4x+12cot2x+logsin x+C1+C2
                        = -14cot4x+12cot2x+logsin x+C     C=C1+C2

Page No 19.69:

Question 12:

cot6 x dx

Answer:

∫ cot6 x dx
= ∫ cot4 x . (cosec2 x – 1) dx
= ∫ cot4 x × cosec2 x dx – â€‹∫ cot4 x dx
= ∫ cot4 x . cosec2 x dx – â€‹∫ cot2 x . cot2 x dx
= ∫ cot4 x – cosec2 x dx – â€‹∫ (cosec2 x – 1) cot2 x dx
= ∫ cot4 x . cosec2 x dx – â€‹∫ cot2 x . cosec2 x dx + â€‹∫ cot2 x dx
= ∫ cot4 x . cosec2 x dx – â€‹∫ cot2 x . cosec2 x dx + â€‹∫ (cosec2 x – 1) dx
Now, let I1= ∫ cot4 x . cosec2 x dx – â€‹∫ cot2 x . cosec2 x dx
And I2= ∫ (cosec2 x – 1) dx
First we integrate I1
I1= ∫ cot4 x . cosec2 x dx – â€‹∫ cot2 x . cosec2 x dx
Let cot x = t
⇒ –cosec2 x dx = dt
⇒ cosec2 dx = – dt
I1=– ∫ â€‹t4 dt + â€‹∫ t2 dt
  =-t55+t33+C1=-cot5 x5+cot3 x3+C1
Now we integrate I2
I2= ∫ (cosec2 x – 1) dx
   =
– cot xx + C1
Now, ∫ cot6 x dx=I1 + I2
                        = -15cot5x+13cot3x-cot x-x+C1+C2
                        = -15cot5x+13cot3x-cot x-x+C      C=C1+C214cot4x+12cot2x+log|sin x|+C     [C=C1+C2]



Page No 19.73:

Question 1:

sin4 x cos3 x dx

Answer:

∫ sin4 x cos3 x dx
=​ ∫ sin4 x . cos2 x cos x dx
= ∫ sin4 x . (1 – sin2 x ) cos x dx
Let sin x = t
⇒​​ cos x dx = dt
Now, ∫ sin4 x . (1 – sin2 x ) cos x dx
= ​​∫ t4 (1 – t2) dt
= ∫ (t4t6) dt
=t55-t77+C=sin5 x5-sin7 x7+C

Page No 19.73:

Question 2:

sin5 x dx

Answer:

∫ sin5 x dx
= ​∫ sin4 x . sin x dx
= ∫ (1 – cos2 x)2 sin x dx
= ∫ (1 – cos4 x – 2 cos2 x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos4 x – 2 cos2 x) sin x dx
=–​∫ (1 + t4 – 2t2) dt
=-t+t55-2t33+C=-t-t55+2t33+C=-cos x+23cos3 x-cos5 x5+C

Page No 19.73:

Question 3:

cos5 x dx

Answer:

∫ cos5 x dx
= ∫ cos4 x . cos x dx
= ∫ (1 – sin2 x)2 cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2 x)2 cos x dx
= ​​∫ (1 – t2)2 . dt
= ∫ (1 + t4 – 2t2) dt
= ∫ dt + â€‹∫ t4 dt – 2 â€‹∫t2 dt
=t+t55-2t33+C=sin x+sin5 x5-23sin3 x+C

Page No 19.73:

Question 4:

sin5 x cos x dx

Answer:

∫ sin5 x cos x dx
Let sin x = t
cos x dx = dt
Now, ∫ sin5 x cos x dx
​= ∫ t5 . dt
=t66+C=sin6 x6+C

Page No 19.73:

Question 5:

sin3 x cos6 x dx

Answer:

∫ sin3 x . cos6 x dx
=​ ∫ sin2 x . cos6 x . sin x dx
= ​​∫ (1 – cos2 x) . cos6 x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ​​∫ (1 – cos2 x) . cos6 x . sin x dx
= –∫ (1 – t2) . t6 dt
= ∫ (t2 – 1) t6 dt
= ∫ (t8t6) dt
=t99-t77+C=cos9 x9-cos7 x7+C

Page No 19.73:

Question 6:

cos7 x dx

Answer:

∫​ cos7 x dx
= ​∫ cos6 x . cos x dx
= ∫ (cos2 x)3 cos x dx
= ∫ (1 – sin2 x)3 . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2 x)3.cos x dx
= ∫ (1 – t2)3 dt
= ∫ (1 – t6 – 3t2 + 3t4) dt
=t-t77-3t33+3t55+C=sin x-17sin7 x-sin3 x+35sin5x+C

Page No 19.73:

Question 7:

x cos3 x2 sin x2 dx

Answer:

x . cos3 x2 sin x2 dx
Let x2 = t
⇒​ 2x dx = dt
x dx=dt2Now, x. cos3x2 sin x2dx=12 cos3 t. sin t . dtAgain let cos t = p-sin t dt = dpsin t dt = -dpSo, 12 cos3 t. sin t . dt =-12p3 dp=-12 p44+C=-p48+C=-cos4 t8+C=-cos4 x28+C

Page No 19.73:

Question 8:

sin7 x dx

Answer:

∫ sin7 x dx
= ∫ sin6 x . sin x dx
= ∫ (sin2 x)3 sin x dx
= ​​∫ (1 – cos2 x)3 sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2 x)3 sin x dx
= ∫ (1 – t2)3 . (–dt)
= –​∫ (1 – t6 – 3t2 + 3t4) dt
=-t-t77-t3+3t55+C=-cos x-cos7 x7-cos3 x+35cos5 x+C=-cos x+17cos7 x+cos3 x-35cos5 x+C

Page No 19.73:

Question 9:

sin3 x cos5 x dx

Answer:

∫ sin3 x . cos5 x dx
​= ∫ sin2 x . cos5 x . sin x dx
= ∫ (1 – cos2 x) . cos5 x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2 x) . cos5 x sin x dx
= –​∫ (1 – t2) t5 dt
= –∫ (t5t7) dt
= ∫(t7t5) dt
=t88-t66+C=cos8 x8-cos6 x6+C

Page No 19.73:

Question 10:

1sin4 x cos2 x dx

Answer:

dxsin4 x.cos2 xDividing numerator & denominator by sin2 x=1sin2 xsin4 x.cot2 xdx=cosec6 xcot2dx=cosec4 x.cosec2 x dxcot2 x=1+cot2 x2.cosec2 x dxcot2 xLet cot x=t-cosec2 x=dtdx-cosec2 x dx=dtNow, 1+cot2 x2.cosec2 x dxcot2 x=1+t2t2 -dt=-1+t4+2t2t2dt=-t-2+t2+2dt=-t-2+1-2+1+t33+2t+C=--1t+t33+2t+C=-13t3-2t+1t+C=-13cot3 x-2 cot x+1cot x+C=-13cot3 x-2 cot x+tan x+C

Page No 19.73:

Question 11:

1sin3 x cos5 x dx

Answer:

dxsin3 x.cos5 xdxDividing numerator & denominator by cos8 x=1cos8 xdxsin3 xcos3 x=sec8 xtan3 xdx=sec6 x.sec2 x dxtan3 x=1+tan2 x3. sec2 x dxtan3 xLet tan x=t sec2 x dx=dtNow, 1+tan2 x3. sec2 x dxtan3 x=1+t23t3.dt=1+t6+3t2+3t4t3dt=1t3+t3+3t+3tdt=t-3 dt+t3 dt+3dtt+3t dt=t-3+1-3+1+t3+13+1+3 log t+3t22+C=-12 tan x-2 +14tan4 x+3 log tan x+32 tan2 x+C

Page No 19.73:

Question 12:

1sin3 x cos x dx

Answer:

dxsin3 x . cos xDividing numerator & denominator by sin4 x=1sin4 xdxsin3 x.cos xsin4 x=cosec4 x dxcot x=cosec2 x.cosec2 x dxcot x=1+cot2 x.cosec2 x dxcot xLet cot x=t-cosec2 x=dtdxcosec2 x dx=-dtNow, 1+cot2 x.cosec2 x cot xdx=1+t2.-dtt=-1t+tdt=-log t-t22+C=-log cot x-cot2 x2+C=log cot x-1-cosec2 x-12+C=log 1cot x-cosec2 x2+12+C=log tan x-12sin2 x+C'      C'=C+12

Page No 19.73:

Question 13:

1sin x cos3 x dx

Answer:

dxsin x. cos3 xDividing numerator & denominaor by cos4 x=1cos4 x dxsin x.cos3 xcos4 x=sec4 x dxtan x=sec2 x . sec2 x dxtan x=1+tan2 x . sec2 x tan xdxLet tan x = tsec2 x = dxdtsec2 x dx =dtNow, 1+tan2 x . sec2 x tan xdx =1+t2 tdt=1t+tdt=log t+t22+C=log tan x+tan2 x2+C



Page No 19.79:

Question 1:

Evaluate the following integrals:

x2a2-x232dx

Answer:

Let I=x2a2-x232dx              Let x=a cosθ             On differentiating both sides,we get              dx=-a sinθ dθ I=a2cos2θa2-a2cos2θ32×-a sinθ dθ      =-a3 cos2θ sinθa31-cos2θ32dθ      =-cos2θ sinθsin3θdθ      =-cot2θ dθ      =-cosec2θ-1 dθ      =--cotθ-θ+c      =cotθ+θ+c      =cotcos-1xa+cos-1xa+c      =cotcot-1xa2-x2+cos-1xa+c      =xa2-x2+cos-1xa+cHence, x2a2-x232dx=xa2-x2+cos-1xa+c

Page No 19.79:

Question 2:

Evaluate the following integrals:

x7a2-x25dx

Answer:

Let I=x7a2-x25dx              Let x=a sinθ             On differentiating both sides,we get              dx=a cosθ dθ I=a8 sin7θ cosθa2-a2sin2θ5dθ      =a8 sin7θ cosθa101-sin2θ5dθ      =sin7θa2 cos9θdθ      =1a2tan7θ sec2θ dθ              Let tanθ=t             On differentiating both sides,we get              sec2θ dθ=dtI =1a2t7 dt      =1a2t88+c      =18a2tan8θ+c      =18a2tansin-1xa8+c      =18a2tantan-1xa2-x28+c      =18a2xa2-x28+c      =18a2x8a2-x24+cHence, x7a2-x25dx=18a2x8a2-x24+c

Page No 19.79:

Question 3:

Evaluate the following integrals:

cos2cot-11+x1-xdx

Answer:

Let I=cos2cot-11+x1-xdx              Let x=cos2θ             On differentiating both sides,we get              dx=-2 sin2θ dθ I=-cos2cot-11+cos2θ1-cos2θ2 sin2θ dθ      =-2cos2cot-12cos2θ2sin2θsin2θ dθ      =-2cos2cot-1cotθsin2θ dθ      =-2cos2θ sin2θ dθ      =-sin4θ dθ      =cos4θ4+c1      =142cos22θ-1+c1      =12x2-14+c1      =12x2+c, where c=-14+c1Hence, cos2cot-11+x1-xdx=12x2+c

Page No 19.79:

Question 4:

Evaluate the following integrals:

1+x2x4dx

Answer:

Let I=1+x2x4dx              Let x=tanθ             On differentiating both sides,we get              dx=sec2θ dθ I=1+tan2θtan4θsec2θ dθ      =sec3θtan4θdθ      =cosθsin4θdθ      =cotθ cosec3θ dθ              Let cosec3θ=t             On differentiating both sides,we get             -3 cosec3θ cotθ dθ=dt I=-13cotθ cosec3θ dtcosec3θ cotθ      =-t3+c      =-13cosec3θ+c      =-13cosectan-1x3+c      =-13coseccosec-11+x2x3+c      =-131+x2x3+cHence, 1+x2x4dx=-131+x2x3+c

Page No 19.79:

Question 5:

Evaluate the following integrals:

1x2+2x+102dx

Answer:

Let I=1x2+2x+102dx       =1x+12+322dx              Let x+1=3tanθ             On differentiating both sides,we get              dx=3sec2θ dθ I=132tan2θ+3223sec2θ dθ      =127sec2θsec4θdθ      =1271sec2θdθ      =127cos2θ dθ      =1541+cos2θ dθ      =154θ+sin2θ2+c      =154θ+tanθ1+tan2θ+c      =154tan-1x+13+tantan-1x+131+tan2tan-1x+13+c      =154tan-1x+13+x+131+x+132+c      =154tan-1x+13+x+13x2+2x+109+c      =154tan-1x+13+3x+1x2+2x+10+cHence, 1x2+2x+102dx=154tan-1x+13+3x+1x2+2x+10+c



Page No 19.83:

Question 1:

1a2-b2 x2 dx

Answer:

dxa2-b2x2= 1b2dxa2b2-x2                                             = 1b2×12ab log ab+xab-x+C      dxa2-x2= 12a log a+xa-x+C= 12ab log a+bxa-bx+c

Page No 19.83:

Question 2:

1a2 x2-b2 dx

Answer:

dxa2x2-b2                                  = 1a2dxx2-ba2= 1a2×12ba log x-bax+ba+C     dxx2-a2=12a log x-ax+a+C= 12ab log ax-bax+b+C

Page No 19.83:

Question 3:

1a2 x2+b2 dx

Answer:

dxa2x2+b2= 1a2dxx2+ba2                = 1a2×abtan-1xba+C   dxa2+x2=1atan-1xa+C= 1abtan-1axb+C

Page No 19.83:

Question 4:

x2-1x2+4 dx

Answer:

x2-1x2+4dx                                         = x2+4-4-1x2+4dx                   = x2+4x2+4dx-5dxx2+22= dx-5dxx2+22= x-52tan-1x2+C    dxx2+a2= 1atan-1xa+C

Page No 19.83:

Question 5:

11+4x2 dx

Answer:

dx1+4x2= dx1+2x2let 2x=t2dx=dtdx=dt2Now, dx1+2x2= 12dt1+t2                     = 12 log t+1+t2+C         dxx2+a2=log x+x2+a2+C= 12 log 2x+1+4x2+C

Page No 19.83:

Question 6:

1a2+b2 x2 dx

Answer:

dxa2+b2x2= dxb2a2b2+x2= 1bdxx2+ab2=1b log x+x2+a2b2+C= 1blog x+b2x2+a2b+C= 1blog bx+b2x2+a2b+C= 1blog bx+b2x2+a2-log b+C= 1b log bx+b2x2+a2-log bb+Clet C-log bb=C'= 1blog bx+b2x2+a2+C'

Page No 19.83:

Question 7:

1a2-b2 x2 dx

Answer:

dxa2-b2x2= dxb2a2b2-x2= 1bdxab2-x2= 1bsin-1xba+C

Page No 19.83:

Question 8:

12-x2+1 dx

Answer:

dx2-x2+1let 2-x=t -dx=dt dx=-dtNow,dx2-x2+1 =-dtt2+1= -log t+t2+1+C= -log 2-x+2-x2+1+C

Page No 19.83:

Question 9:

12-x2-1 dx

Answer:

dx2-x2-1let 2-x=t -dx=dt dx=-dtNow, dx2-x2-1= -dtt2-1= -log t+t2-1+C= -log 2-x+2-x2-1+C

Page No 19.83:

Question 10:

x4+1x2+1 dx

Answer:

x4+1x2+1dx= x4-1+1+1x2+1dx= x4-1x2+1+2x2+1dx= x2-1x2+1x2+1+2x2+1dx= x2-1+2x2+1dx= x33-x+2tan-1x+C



Page No 19.86:

Question 1:

14x2+12x+5 dx

Answer:

dx4x2+12x+5= 14dxx2+3x+54= 14dxx2+3x+322-322+54= 14dxx+322-94+54= 14dxx+322-12let x+32=tdx=dtNow, 14dxx+322-12= 14dxt2-12= 14×12×1 log t-1t+1+C= 18 log x+32-1x+32+1+C= 18 log x+12x+52+C= 18 log 2x+12x+5+C

Page No 19.86:

Question 2:

1x2-10x+34 dx

Answer:

dxx2-10x+34= dxx2-10x+25-25+34= dxx-52+9= dxx-52+32let x-5=t dx=dtNow,  dxx-52+32= dtt2+32= 13tan-1t3+C= 13tan-1x-53+C

Page No 19.86:

Question 3:

11+x-x2 dx

Answer:

dx1+x-x2= -dxx2-x-1= -dxx2-x+14-14-1= -dxx-122-54= dx54-x-122= dx522-x-122let x-12=t dx=dtNow,  dx522-x-122= dt522-t2= 12×52 log 52+t52-t+C
= 15 log 5+2t5-2t+C= 15 log 5+2x-125-2x-12+C= 15 log 5-1+2x5+1-2x+C

Page No 19.86:

Question 4:

12x2-x-1 dx

Answer:

dx2x2-x-1= 12dxx2-x2-12= 12dxx2-x2+142-142-12= 12dxx-142-116-12= 12dxx-142-1+816= 12dxx-142-342let x-14=t dx=dt
Now, 12dxx-122-342= 12dtt2-342= 12dtt2-342= 12×34×12 log t-34t+34+C= 23×12 log x-14-34x-14+34+C= 23×12 log x-1x+12+C= 13 log 2x-12x+1+C= 13 log x-12x+1+log2+C= 13 log x-12x+1+13 log 2+C= 13 log x-12x+1+C'         C'=13 log 2+C

Page No 19.86:

Question 5:

1x2+6x+13 dx

Answer:

dxx2+6x+13= dxx2+2×x×3+9-9+13= dxx+32+22= 12 tan-1x+32+C



Page No 19.90:

Question 1:

sec2 x1-tan2 x dx

Answer:

sec2 x dx1-tan2 xlet tan x=t sec2 x dx=dtNow, sec2 x dx1-tan2 x= dt1-t2= 12 log 1+t1-t+C= 12 log 1+tanx1-tanx+C

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Question 2:

ex1+e2x dx

Answer:

exdx1+e2xlet ex=texdx=dtNow, exdx1+e2x=dt1+t2= tan-1t+C= tan-1ex+C

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Question 3:

cos xsin2 x+4 sin x+5 dx

Answer:

cos x dxsin2 x+4sin x+5let sin x =t cos x dx=dtNow,cos x dxsin2 x+4sin x+5 = dtt2+4t+5= dtt2+2×t×2+4+1= dtt+22+12= 11 tan-1t+21+C= tan-1sin x+2+C

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Question 4:

exe2x+5 ex+6 dx

Answer:

ex dxe2x+5ex+6let ex=t ex dx=dtNow, ex dxe2x+5ex+6=dtt2+5t+6= dtt2+5t+522-522+6= dtt+522-254+6= dtt+522-25+244= dtt+522-122= 12×12 log t+52-12t+52+12+C= log t+2t+3+C= log ex+2ex+3+C

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Question 5:

e3x4 e6x-9 dx

Answer:

e3x dx4e6x-9let e3x=t e3x×3dx=dt e3x dx=dt3Now, e3x dx4e6x-9= 13dt4t2-9= 13dt2t2-32= 13×12×3 log 2t-32t+3×12+C= 136 log 2t-32t+3+C= 136 log 2e3x-32e3x+3+C

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Question 6:

dxex+e-x

Answer:

dxex+e-x= dxex+1ex= ex dxe2x +1let ex=t ex dx=dtNow, ex dxe2x+1= dt1+t2= tan-1t+c= tan-1ex+c

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Question 7:

xx4+2x2+3 dx

Answer:

x dxx4+2x2+3let x2=t 2x dx=dt x dx =dt2Now, x dxx4+2x2+3= 12dtt2+2t+3= 12dtt2+2t+1+2= 12dtt+12+22                                        = 12×12 tan-1t+12+C                   dxx2+a2=1atan-1xa+C= 122 tan-1x2+12+C

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Question 8:

3x51+x12 dx

Answer:

3x51+x12dxlet x6=t 6x5 dx=dt x5 dx=dt6Now, 3x51+x12dx= 36dt1+t2= 12 tan-1t+C

= 12 tan-1x6+C

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Question 9:

x2 dxx6-a6 dx

Answer:

x2dxx6-a6let x3=t 3x2 dx=dt x2 dx=dt3Now, x2dxx6-a6= 13dtt2-a32= 13×12a3 log t-a3t+a3+C= 16a3 log x3-a3x3+a3+C

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Question 10:

x2x6+a6 dx

Answer:

x2x6+a6dx x2dxx32+a32let x3=t 3x2dx=dt x2dx=dt3Now, x2x6+a6dx= 13dtt2+a32= 13a3tan-1ta3+C= 13a3tan-1x3a3+C

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Question 11:

1x x6+1 dx

Answer:

dxxx6+1= x5dxx6x6+1let x6=t 6x5dx=dt x5dx=dt6Now, dxx6x6+1= 16dttt+1= 16dtt2+t= 16dtt2+t+14-14= 16dtt+122-122= 16×12×12 log t+12-12t+12+12+C= 16 log tt+1+C= 16 log x6x6+1+C

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Question 12:

xx4-x2+1 dx

Answer:

x dxx4-x2+1Let x2=t 2x dx=dt x dx=dt2Now, x dxx4-x2+1=12dtt2-t+1= 12dtt2-t+122-122+1= 12dtt-122+34= 12dtt-122+322= 12×23tan-1t-1232+C= 13tan-12t-13+C= 13tan-12x2-13+C

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Question 13:

x3x4-18x2+11 dx

Answer:

x dx3x4-18x2+11let x2=t 2x dx=dt x dx=dt2Now, x dx3x4-18x2+11= 12dt3t2-18t+11= 13×2dtt2-6t+113= 16dtt2-6t+9-9+113= 16dtt-32-163= 16dtt-32-432= 16×12×43 log t-3-43t-3+43+C= 348 log x2-3-43x2-3+43+C

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Question 14:

ex1+ex2+ex dx

Answer:

exdx1+ex2+exlet ex=t= exdx=dt= dt1+t2+t= dt2+t+2t+t2= dtt2+3t+2= dtt2+3t+322-322+2= dtt+322-94+2= dtt+322-14= dtt+322-122

= 12×12 log t+32-12t+32+12+C= log t+1t+2+C= log ex+1ex+2+C

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Question 15:

Evaluate the following integrals:

1cosx+cosecxdx

Answer:

Let I=1cosx+cosecxdx       =sinx1+cosx sinxdx       =2sinx2+2cosx sinxdx       =sinx+cosx+sinx-cosx2+2cosx sinxdx       =sinx+cosx2+2cosx sinxdx+sinx-cosx2+2cosx sinxdx       =sinx+cosx3-sin2x-cos2x+2cosx sinxdx+sinx-cosx1+sin2x+cos2x+2cosx sinxdx       =sinx+cosx3-sinx-cosx2dx+sinx-cosx1+sinx+cosx2dx       = I1+I2            ...(1)          where, I1=sinx+cosx3-sinx-cosx2dx and I2=sinx-cosx1+sinx+cosx2dxNow,I1=sinx+cosx3-sinx-cosx2dx                       Let sinx-cosx=t                       On differentiating both sides, we get                       cosx+sinxdx=dtI1=13-t2dt      =123log3+t3-t+c1      =123log3+sinx-cosx3-sinx-cosx+c1      ...(2)Now,I2=sinx-cosx1+sinx+cosx2dx                       Let sinx+cosx=t                       On differentiating both sides, we get                       cosx-sinxdx=dtI2=-11+t2dt      =-tan-1t+c2      =-tan-1sinx+cosx+c2      ...(3)On substituting (2) and (3) in (1), we getI=123log3+sinx-cosx3-sinx+cosx-tan-1sinx+cosx+cHence, 1cosx+cosecxdx=123log3+sinx-cosx3-sinx+cosx-tan-1sinx+cosx+c



Page No 19.93:

Question 1:

12x-x2 dx

Answer:

dx2x-x2= dx2x-x2-1+1= dx1-x2-2x+1= dx1-x-12                                = sin-1x-1+C        dxa2-x2=sin-1xa+C

Page No 19.93:

Question 2:

18+3x-x2 dx

Answer:

dx8+3x-x2 dx8-x2-3x dx8-x2-3x+322-322 dx8-x-322+94 dx4122-x-322 sin-1x-32412+C sin-12x-341+C

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Question 3:

15-4x-2x2 dx

Answer:

dx5-4x-2x2= dx252-2x-x2= 12dx52-2x-x2= 12dx52-x2+2x= 12dx52-x2+2x+1-1= 12dx52-x+12+1= 12dx72-x+12= 12dx722-x+12= 12sin-1x+127+C= 12sin-127x+1+C

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Question 4:

13x2+5x+7 dx

Answer:

     dx3x2+5x+7=dx3x2+53x+73=13dxx2+53x+562-562+73=13dxx+562-2536+73=13dxx+562+-25+8436=13dxx+562+5936=13dxx+562+59362=13 log x+56+x+562+5936+C=13 log x+56+x2+53x+73+C

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Question 5:

1x-αβ-x dx, β>α

Answer:

  Let I=dxx-α β-x=dxβx-x2-αβ+αx=dx-x2+α+β x-αβ=dx-x2-α+β x+αβ=dx-x2-α+β x+α+β22-α+β22+αβ=dx-x-α+β22+α+β22-αβ=dx-x-α+β22+α+β2-4αβ4=dx-x-α+β22+α-β22=dxα-β22-x-α+β22=sin-1 x-α+β2α-β2+C=sin-1 2x-α-βα-β+C

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Question 6:

17-3x-2x2 dx

Answer:

     dx7-3x-2x2=12dx72-32x-x2=12dx72-x2-32x=12dx722-x2+32x+342-342=12dx722-x+342+916=12dx72+916-x+342=12dx56+916-x+342=12dx6542-x+342=12 sin-1 x+34654+C=12sin-1 4x+365+C

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Question 7:

116-6x-x2 dx

Answer:

     dx16-6x-x2=dx16-x2+6x=dx16-x2+6x+32-32=dx16+9-x+32=dx52-x+32=sin-1 x+35+C

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Question 8:

17-6x-x2 dx

Answer:

     dx7-6x-x2=dx7-x2+6x=dx7-x2+6x+32-32=dx7+9-x+32=dx42-x+32=sin-1x+34+C

Page No 19.93:

Question 9:

15x2-2x dx

Answer:

     dx5x2-2x=dx5x2-25x=15dxx2-25x+152-152=15dxx-152-152=15 log x-15+x-152+152+C=15 log 5x-15+5x2-2x5+C



Page No 19.98:

Question 1:

xx4+a4 dx

Answer:

     x dxx4+a4=x dxx22+a22let x2=t2x dx=dtx dx=dt2Now, x dxx22+a22=12dtt2+a22=12 log t+t2+a4+C=12 log x2+x4+a4+C

Page No 19.98:

Question 2:

sec2 x4+tan2 x dx

Answer:

sec2x dx4+tan2 xlet tan x=tsec2x dx=dtNow, sec2x dx4+tan2 x=dt22+t2=log t+4+t2+C=log tan x+4+tan2x+C

Page No 19.98:

Question 3:

ex16-e2x dx

Answer:

     ex dx16-ex2let ex=tex dx=dtNow, ex dx16-ex2= dt16-t2=dt42-t2=sin-1 t4+C=sin-1 ex4+C



Page No 19.99:

Question 4:

cos x4+sin2 x dx

Answer:

cos x dx4+sin2xlet sin x=tcos x dx=dtNow,cos x dx4+sin2x =dt22+t2=log t+4+t2+C=log sin x+4+sin2x+C

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Question 5:

sin x4 cos2 x-1 dx

Answer:

sin x dx4 cos2x-1let cos x=t-sin x dx=dtsin x dx=-dtNow, sin x dx4 cos2x-1=-dt4t2-1=-dt4t2-14=-1 2dtt2-122=-12 log t+t2-14+C=-1 2 log t+4t2-12+C=-12 log 2t+4t2-12+C=-12log 2t+4t2-1-log 2+C=-12 log 2t+4t2-1+log 22+Clet C´=log 22+C=-12 log 2cost+4cos2t-1+C´

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Question 6:

x4-x4 dx

Answer:

     x dx4-x4x dx22-x22let x2=t2x dx=dtx dx=dt2Now, x dx22-x22=12dt22-t2=12×sin-1 12+C=12sin-1 x22+C

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Question 7:

1x4-9 log x2 dx

Answer:

dxx4-9 log x2let log x=t1x dx=dtNow, dxx4-9 log x2=dt4-9t2=dt22-3t2=13 sin-1 3t2+C=13 sin-1 3 log x2+C

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Question 8:

sin 8x9+sin4 4x dx

Answer:

     sin 8x dx9+sin4 4x2 sin 4x·cos 4 x9+sin24x2dxlet sin2 4x=t2 sin 4x·cos 4x × 4 dx=dt2 sin 4x cos 4x dx=dt4Now, 2 sin 4x·cos 4 x9+sin24x2dx=14dt9+t2=14dt32+t2=14 log t+32+t2+C=14 log sin2 4x+9+sin4 4x+C

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Question 9:

cos 2xsin2 2x+8 dx

Answer:

cos 2 x·dxsin2 2x+8let sin 2x=tcos 2x×2·dx=dtcos 2x·dx=dt2Now, cos 2 x·dxsin2 2x+8 =12dtt2+222=12log t+t2+8+C=12 log sin 2x+sin2 2x+8+C

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Question 10:

sin 2xsin4 x+4 sin2 x-2 dx

Answer:

sin 2 x dxsin4x+4 sin2x-2let sin2x=t2 sin x cos x dx=dtsin 2 x dx=dtNow, sin 2 x dxsin4x+4 sin2x-2=dtt2+4t-2=dtt2+4t+4-4-2=dtt+22-62=log t+2+t+22-6+C=log t+2+t2+4t-2+C=log sin2x+2+sin4x+4 sin2x-2+C

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Question 11:

sin 2xcos4 x-sin2 x+2 dx

Answer:

sin 2 x dxcos4x-sin2 x+22 sin x cos x dxcos4x-1-cos2x+22 sin x cos xcos4x+cos2x+1Let cos2x =t2 cos x ×-sin x dx=dtsin 2x dx=-dtNow, sin 2 x dxcos4x-sin2 x+2=-dtt2+t+1=-dtt2+t+122-122+1=-dtt+122+34=-dtt+122+322=-log t+12+t+122+322+C=-log t+12+t2+t+1+C=-log cos2x+12+cos4x+cos2x+1+C

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Question 12:

cos x4-sin2 x dx

Answer:

cos x dx4-sin2xlet sin x=tcos x dx=dtNow, cos x dx4-sin2x=dt4-t2=dt22-t2=sin-1 t2+C=sin-1 sin x2+C

Page No 19.99:

Question 13:

1x2/3x2/3-4 dx

Answer:

dxx23x23-22=dxx23x132-22Let x13=t13 x-23 dx=dt13x23 dx=dtdxx23=3 dtNow, dxx23x23-22=3dtt2-22=3 log t+t2-22+C=3 log x13+x23-4+C

Page No 19.99:

Question 14:

11-x29+sin-1 x2 dx

Answer:

dx1-x2 9+sin-1 x 2let sin-1x=t11-x2 dx=dtNow, dx1-x2 9+sin-1 x 2 =dt9+t2=dt32+t2=log t+32+t2+C=log sin-1x+9+sin-1x2+C

Page No 19.99:

Question 15:

cos xsin2 x-2 sin x-3 dx

Answer:

cos x dxsin2x-2 sin x-3let sin x=tcos x dx=dtNow, cos x dxsin2x-2 sin x-3=dtt2-2t-3=dtt2-2t+1-1-3=dtt-12-22=log t-1+t-12-22+C=log t-1+t2-2t-3+C=log sin x-1+sin2x-2 sin x-3+C

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Question 16:

cosec x-1 dx

Answer:

cosec x-1  dx=1sin x-1dx=1-sin xsin xdx=1-sin x 1+sin xsin x 1+sin xdx=1-sin2xsin2x+sinxdx=cos x dxsin2x+sin xLet sin x=tcos x dx=dtNow, cos x dxsin2x+sin x=dtt2+t     dtt2+t=dtt2+t+122- 122=dtt+122-122=log t+12+t+122-122+C=log t+12+t2+t+C=log sin x+12+sin2x+sinx+C

Page No 19.99:

Question 17:

sin x-cos xsin 2x dx

Answer:

sin x-cosxsin 2x dx=sin x-cos x1+sin 2x-1dx=sin x-cos xsin2x+cos2x+2 sin x cos x-1dx=sin x-cos xsin x+cos x2-1dxlet sin x+cos x=tcos x-sin x dx=dtsin x-cos xdx=-dtNow, sin x-cos xsin x+cos x2-1dx=-dtt2-12=-log t+t2-1+C=-log sin x+cos x+sin x+cos x2-1+C=-log sin x+cos x+sin2 x+cos2 x +2sinx.cosx-1+C=-log sin x+cos x+sin 2x+C

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Question 18:

cosx-sinx8-sin2xdx

Answer:

Let I=cosx-sinx8-sin2xdx       =cosx-sinx9-1-sin2xdx       =cosx-sinx9-sin2x-cos2x-2sinxcosxdx       =cosx-sinx9-sinx+cosx2dx                       Let sinx+cosx=t                       On differentiating both sides, we get                       cosx-sinxdx=dtI=132-t2dt     =sin-1t3+c     =sin-1sinx+cosx3+cHence, cosx-sinx8-sin2xdx=sin-1sinx+cosx3+c



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