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 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 (