RD Sharma XII Vol 1 2018 Solutions for Class 12 Science 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 Science 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 2018 Book of class 12 Science 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 2018 Solutions. All RD Sharma XII Vol 1 2018 Solutions for class 12 Science 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
= ​∫ 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.dθ
& θ = sin–1 x
sin–1 (3x – 4x3)dx = sin–1 (3 sin ​θ – 4 sin3 ​θ) . cos ​θ d​θ
                                = ∫ sin–1 (sin 3​θ) . cos ​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