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NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

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Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 7 - Integrals solved by expert Maths teachers on Vedantu.com as per NCERT (Central Board of Secondary Education) Book guidelines. All Chapter 7 - Integrals Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.


NCERT Exemplar for Class 12 Maths, Chapter 7 - Integrals is available on Vedantu and its solutions are available on Vedantu in PDF form. The PDF of Chapter 7 - Integrals consists of important topics and additional questions.

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Access NCERT Exemplar Solutions for CBSE Class 12 Mathematics Chapter 7 - Integrals

Solved Examples

Short Answer (S.A.)

1. Integrate (2axbx2+3cx23) w.r.t. x.

Ans: let I= (2axbx2+3cx23)dx

I=2ax12dxbx2dx+3cx23 dx

I=2a×2×x+bx+95 c x53+C

I=4ax+bx+95 c x53+C

Hence (2axbx2+3cx23)dx=4ax+bx+95 c x53+C.


2. Evaluate 3axb2+c2x2 dx

Ans: let I=3axb2+c2x2 dx

Now, put b2+c2x2=t

2c2x dx=dt

x dx=dt2c2

3ax dx=3a dt2c2

Therefore, 

I=3a 2c2dtt 

I=3a 2c2 log|t|+C

I=3a 2c2 log|b2+c2x2|+C

Hence, 3axb2+c2x2 dx=3a 2c2 log|b2+c2x2|+C.


3. Verify the following using the concept of integration as an antiderivative.

x3x+1dx=xx22+x33log|x+1|+C 

Ans: here, L.H.S =xx22+x33log|x+1|+C 

Now, differentiate w.r.t x , we get

⇒ L.H.S =ddx(xx22+x33log|x+1|+C)

=12x2+3x231(x+1)+0

=1x+x21(x+1)

 =(x+1)x(x+1)+x2(x+1)1(x+1)

=x+1x2x+x3+x21(x+1)

=x3(x+1)

Therefore, (xx22+x33log|x+1|+C)=x3(x+1) dx

Hence verified.


4. Evaluate 1+x1x dx, x1.

Ans: Let I=1+x1x dx

I=(1+x)(1+x)(1x)(1+x) dx

I=(1+x)21x2 dx

I=1+x1x2 dx

I=11x2 dx+x1x2 dx

I=sin1x+x1x2 dx

Now, put 1x2=t2

2x dx=2t dt

x dx=t dt

Therefore, 

I=sin1xt dtt 

I=sin1x dt

I=sin1xt+C

I=sin1x1x2+C

Hence, 1+x1x dx=sin1x1x2+C


5. Evaluate dx(xα)(βx), β>α

Ans: Let I=dx(xα)(βx)

Now, put xα=t2

dx=2t dt

And βx=β(t2+α)=βαt2

Therefore, 

I=2t dtt2(βαt2)

I=2 dt(βαt2)

I=2 dt[(βα)2t2]

I=2 sin1tβα+C

I=2 sin1xαβα+C

Hence, dx(xα)(βx)=2 sin1xαβα+C.


6. Evaluate tan8x sec4x dx

Ans: let I=tan8x sec4x dx

I=tan8x (sec2x) sec2x dx

I=tan8x (1+tan2x) sec2x dx

Now, put tanx=t

sec2x dx=dt

Therefore, 

I=t8 (1+t2) dt

I=(t8+t10)  dt

I=t99+t1111+C

I=tan9x9+tan11x11+C

Hence, tan8x sec4x dx=tan9x9+tan11x11+C


7. Find x3x4+3x2+2dx

Ans: Let I=x3x4+3x2+2dx

I=x2.x dxx4+3x2+2

Now, put x2=t

2x dx=dt

x dx=dt2 

Therefore, 

I=12t dtt2+3t+2

I=12t dt(t+1)(t+2)

Consider t (t+1)(t+2)=At+1+Bt+2

t (t+1)(t+2)=At+2A+Bt+B(t+1)(t+2)

t (t+1)(t+2)=(A+B)t+2A+B(t+1)(t+2)

t=(A+B)t+2A+B

On comparing both sides, we get

A+B=1 and 2A+B=0

On solving, we get

A=1, B=2

Therefore, 

I=12t dt(t+1)(t+2)

I=12[1t+1+2t+2]dt

I=12 (log|t+1|+2log|t+2|)+C

I=12log|t+1|+log|t+2|+C

I=log|t+1|+log|t+2|+C

I=log|t+2t+1|+C

I=log|x2+2x2+1|+C

Hence, I=x3x4+3x2+2dx=log|x2+2x2+1|+C


8. Find dx2 sin2x+ 5cos2x

Ans: Let I=dx2 sin2x+ 5cos2x

Dividing numerator and denominator by cos2x, we get

I=sec2xdx2 tan2x+ 5

Put tanx=t

sec2x dx=dt

Therefore,

I=dt2 t2+ 5

I=12dt t2+52

I=12dt t2+(52)2

I=12×25  tan12t5+C

I=110 tan1(2 tan x5)+C

Hence, dx2 sin2x+ 5cos2x=110 tan1(2 tan x5)+C.


9. Evaluate 21(7x5)dx as a limit of sums.

Ans: here, we have a=1 , b=2 and h=2+1nnh=3

And f(x)=7x5

Now, we have 

21(7x5)dx=limh0h[f(1)+f(1+h)+f(1+2h)+f(1+(n1)h]

Also we have, 

f(1)=75=12

f(1+h)=7+7h5=12+7h

f(1+(n1)h)=7(n1)h12=

Therefore, 

21(7x5)dx=limh0h[12+(7h12)+(14h12)+(7(n1)h12]

21(7x5)dx=limh0h[7h[1+2++(n10)]12n]

21(7x5)dx=limh0h[7h×n(n1)212n]

21(7x5)dx=limh0h[72(nh)(nhh)12nh]

21(7x5)dx=72×3×312×3

21(7x5)dx=63236

21(7x5)dx=63722

21(7x5)dx=92

Hence, 21(7x5)dx=92.


10. Evaluate π20tan7xcot7x+ tan7xdx

Ans: Let I=π20tan7xcot7x+ tan7xdx …………. (i)

I=π20tan7(π2x)cot7(π2x)+ tan7(π2x)dx

I=π20cot7xtan7x+ cot7xdx …………….. (ii)

Now, adding eq (i) and (ii), we get 

2I=π20tan7x+ cot7xtan7x+ cot7xdx

2I=π20dx

2I=[x]0π2

2I=π2

I=π4

Hence, π20tan7xcot7x+ tan7xdx=π4.


11. Find 8210xx+10x dx

Ans: Let I=8210xx+10x dx ………….. (i)

I=828+2(10x)8+2x+8+2 (10x) dx

I=821010+x10x+10 10+x dx

I=82x10x+x dx ……………… (ii)

Now, adding eq (i) and (ii), we get

2I=8210x +x10x+x dx

2I=82dx

2I=[x]28

2I=82

2I=6

I=3

Hence, 8210xx+10x dx=3.


12. Find π401+sin2x  dx

Ans: Let I=π401+sin2x  dx

I=π40sin2x+cos2x+2 sinx . cos x  dx

I=π40(sinx+cosx)2  dx

I=π40(sinx+cosx) dx

I= [cosx+sinx]0π4

I=[cosπ4+sinπ4(cos0+sin0)] 

I=[12+12(1+0)]

I=1

Hence, π401+sin2x  dx=1.


13. Find x2tan1x dx

Ans: Let I=x2tan1x dx

I=tan1xx2 dx(ddx(tan1x)x2 dx)dx

I=x33tan1x11+ x2.x33 dx

I=x33tan1x13x31+ x2 dx

I=x33tan1x13(xx1+ x2) dx

I=x33tan1x13x dx+13x1+ x2 dx 

I=x33tan1xx26+13x1+ x2 dx

Now, put 1+ x2=t

2x dx=dt

x dx=dt2

I=x33tan1xx26+16dtt 

I=x33tan1xx26+16log|t|+C

I=x33tan1xx26+16log|1+ x2|+C


14. Find 104x+4x2  dx

Ans: Let I=104x+4x2  dx

I=4x24x+10  dx

I=4(x2x+104)  dx

I=2[x2x+(12)2(12)2+104]  dx

I=2[(x12)214+104]  dx

I=2[(x12)2+94]  dx

I=2[(x12)2+(32)2]  dx

I= [(x12)24x24x+10 +(32)22log|(x12)+4x24x+10 |]

I=(2x14)4x24x+10 +94log|(x12)+4x24x+10 |


Long Answer (L.A.)

15. Evaluate x2 dxx4+x22

Ans: Let I=x2 dxx4+x22

Now, put x2=t

Therefore, 

x2 x4+x22=t t2+t2=t (t+2)(t1)

t (t+2)(t1)=At+2 +Bt1 

t (t+2)(t1)=AtA+Bt+2B(t+2)(t1) 

t=(A+B)t+2BA

On comparing both side, we get

A+B=1 and 2BA=0

On solving these equations, we get

B=13  and A=23

So, x2 x4+x22=231x2+2+131x21

Therefore, 

I=(231x2+2+131x21)dx

I=231x2+(2)2dx+ 131x21dx

I=23 ×12 tan1x2+13 ×12log|x1x+1|+C

I=23  tan1x2+16 log|x1x+1|+C

Hence, x2 dxx4+x22=23  tan1x2+16 log|x1x+1|+C.


16. Evaluate x3+xx49  dx

Ans: Let I=x3+xx49  dx

I=x3x49  dx+xx49  dx

I=I1+I2 ………….. (i)

where I1=x3x49  dx and I2=xx49  dx

Now, I1=x3x49  dx

Put x49=t

4x3 dx=dt

x3 dx=dt4

Therefore, 

I1=14dtt  

I1=14log|t|+C1

I1=14log|x49|+C1 ……….. (ii)

Now, I2=xx49  dx

Put x2=t

2x dx=dt

x dx=dt2

Therefore, 

I2=12dtt232  

I2=12×12×3log|t3t+3|+C2

I2=112log|x23x2+3|+C2

Now, put the value of I1 and I2 in eq (i), we get 

⇒  I=14log|x49|+C1+112log|x23x2+3|+C2

⇒  I=14log|x49|+112log|x23x2+3|+C


17. Show that π20sin2x dxsinx+cosx=12log(2+1)

Ans: let I=π20sin2x dxsinx+cosx  ………. (i)

I=π20sin2(π2x) dxsin(π2x)+cos(π2x)

I=π20cos2x dxcosx+sinx …………. (ii)

Adding eq (i) and (ii), we get 

2I=π20(sin2x+cos2x) dxcosx+sinx

2I=π20 dxcosx+sinx

2I=12π20 dxcosx 12+sinx. 12

2I=12π20 dxcosx cos π4 +sinx. sin   π4 

2I=12π20 dxcos(x π4) 

2I=12π20sec(x π4) dx 

2I=12[log{sec(x π4)+tan(x π4)   }]0π2

2I=12[log{sec(π2 π4)+tan(π2 π4)   }log{sec( π4)+tan( π4)   }]

2I=12[log{sec( π4)+tan( π4)   }log{sec( π4)tan( π4)   }]

2I=12[log(2+1)log(21)]

2I=12 ×log|2+121|

2I=12 ×log|(2+1)221|

2I=12 ×2log(2+1)

I=12 log(2+1)

Hence proved 


18. Find 10x(tan1x)2 dx

Ans: Let I=10x(tan1x)2 dx

Applying integrating by parts, we get

I=[x22(tan1x)2]01 102 tan1x1 + x2. x22dx 

I=[12(tan11)20] 10 tan1x1 + x2x2dx

I=[12(π4)2] 10 (x2+11)tan1x1 + x2dx

I=π232 10 (x2+11)tan1x1 + x2dx

I=π232 10 (x2+1)tan1x1 + x2dx+10 tan1x1 + x2dx 

I=π232 10tan1x dx+10 tan1x1 + x2dx 

I=π232 I1+I2 ……….. (i)

Where, I1=10tan1x dx and I2=10 tan1x1 + x2dx

Now, I1=10tan1x dx

I1=[x tan1x]01 10x1+x2 dx

I1=[1. tan110] 10x1+x2 dx

I1=π4 10x1+x2 dx

Now, put 1+x2=t

2x dx=dt

x dx=dt2

And at x=0 t=1 

⇒ at x=1 t=2

Therefore, 

I1=π41221dtt 

I1=π412[logt]12

I1=π412[log2log1]

I1=π412log2 ………….. (ii)

Now, I2=10 tan1x1 + x2dx

Put tan1x=t

11+x2 dx=dt

And at x=0t=0

⇒ at x=1t=π4 

Therefore, 

I2=π40t dt

I2=[t22]0π4

I2=12[t2]0π4

I2=12[π2160]

I2=π232

Now, put the value of I2 and I2 in eq(i), we get

I=π232 π4+12log2+π232

I=π216 π4+12log2

Hence, 10x(tan1x)2 dx=π216 π4+12log2.


19. Evaluate 21f(x)dx, where f(x)= |x+1|+|x|+|x1|.

Ans: here, we can redefine f(x) as 

f\left( x \right)=\left\{ 2x,    if1<x0x+2,   if  0<x13x,           if 1<x2 \right.$

Therefore, 21f(x)dx= 01(2x) dx+10(x+2)dx+213x dx

21f(x)dx= [2xx22]10+[x22+2x]01+3[x22]12

21f(x)dx= [0(212)]+[(12+2)0]+3[212]

21f(x)dx=52+52+92

21f(x)dx=192.


Objective Type Questions 

Choose the correct answer from the given four option in each of the Examples from 20 to 30.

20. ex(cosxsinx) dx is equal to

(A) excosx+C                               

(B) exsinx+C         

(C) excosx+C                           

(D) exsinx+C       

Ans: here, as we know that ex[f(x)+f(x)] dx=ex f(x)+C 

Now, if f(x)=cosx, then f(x)=sinx

Therefore, ex(cosxsinx) dx=excosx+C                               

Hence, option (A) is the correct answer.


21. dxsin2x  cos2x is equal to

(A)  tanx+cotx+C                       

(B) (tanx+cotx)2+C       

(C)   tanxcotx+C                      

(D)  (tanxcotx)2+C       

Ans: Let I=dxsin2x  cos2x

I=(sin2x + cos2x) dxsin2x  cos2x

I=sin2x dxsin2x  cos2x+cos2x dxsin2x  cos2x

I= dx  cos2x+ dxsin2x  

I=sec2x dx+cosec2x dx

I=tanxcotx+C

Hence, option (C) is the correct answer.


22. If 3ex5ex4ex+5ex dx=ax+blog|4ex+5ex|+C, then 

(A)  a=18 , b=78                    

(B) a=18 , b=78

(C) a=18 , b=78                  

(D) a=18 , b=78

Ans: here, we have 3ex5ex4ex+5ex dx=ax+blog|4ex+5ex|+C

Now, differentiating both sides, we have 

ddx(3ex5ex4ex+5ex dx)=ddx(ax+blog|4ex+5ex|+C)

3ex5ex4ex+5ex=a+b×14ex+5ex×(4ex5ex)

3ex5ex4ex+5ex=a+b(4ex5ex4ex+5ex)

(3ex5ex)=a(4ex+5ex)+b(4ex5ex)

(3ex5ex)=4aex+5aex+4bex5bex

3ex5ex=4ex(a+b)5ex(a+b)

Now, comparing both sides, we get

4(a+b)=3 and (a+b)=1

(a+b)=34 …… (i)  and (a+b)=1 ………(ii)

Adding eq (i) and (ii), we get 

2b=74

b=78

Now, put the value of b in eq (ii), 

 a+78=1

 a=178

 a=18

Here, a=18 and b=78

Hence, option (A) is the correct answer.


23: b+ca+cf(x)dx is equal to 

(A)  baf(xc)dx                           

(B) baf(x+c)dx

(C) baf(x)dx                                    

(D) bcacf(x)dx

Ans: Let I=b+ca+cf(x)dx

Now, put x=t+c

dx=dt

And at x=a+ct=a

⇒ at x=b+ct=b

Therefore, 

I=baf(t+c)dt

I=baf(x+c)dx

Hence, option (B) is the correct answer.


24. If f and g are continuous function in [0, 1] satisfying f(x)=f(ax) and g(x)+g(ax)=a, then a0f(x).g(x)dx is equal to

(A) a2                                                           

(B) a2 a0f(x) dx

(C) a0f(x) dx                                         

(D) aa0f(x) dx

Ans: Let I=a0f(x).g(x)dx

I=a0f(xa).g(xa)dx

I=a0f(x).(ag(x))dx

I=a0[a f(x)f(x).g(x)]dx

I=aa0 f(x)dxa0f(x).g(x)dx

I=aa0 f(x)dxI

2I=aa0 f(x)dx

I=a2a0 f(x)dx

Hence, option (B) is the correct answer.


25: If x=y0dt1+9t2 and d2ydx2=ay, then a is equal to 

(A)  3                                                

(B) 6

(C) 9  

(D) 1

Ans: here, we have x=y0dt1+9t2

Differentiate w. r.t y, we get 

dxdy=11+9y2

dydx=1+9y2

Now, differentiate w. r.t x , we get 

d2ydx2=18 y21+9y2.dydx

d2ydx2=9 y1+9y2.1+9y2

d2ydx2=9y

Thus, value of x is 9.

Hence, option (C) is the correct answer.


26. 11x3+|x|+1x2+2|x|+1dx is equal to 

(A) log2                                             

(B) 2 log2

(C) 12 log2                                          

(D) 4 log2

Ans: Let I=11x3+|x|+1x2+2|x|+1dx

I=11x3x2+2|x|+1dx+11|x|+1x2+2|x|+1dx

Here, 11x3x2+2|x|+1dx is odd function and 11|x|+1x2+2|x|+1dx is even function.

I=0+210|x|+1(|x|+1)2dx

I=210x +1(x +1)2dx

I=2101(x +1)dx

I=2[log|x+1|]01

I=2[log2log1]

I=2log2

Hence, option (B) is the correct answer.


27. If 10et1+ t dt=a, then 10et(1+ t)2 dt is equal to 

(A) a1+e2                             

(B) a+1e2

(C) a1e2                             

(D) a+1+e2

Ans: here, we have 10et1+ t dt=a

a=10et1+ t dt

Applying integrating by parts, we get

a=[et×1(1+ t)]01+10et(1+ t)2 dt

a=[e21]+10et(1+ t)2 dt

10et(1+ t)2 dt=a+1e2

Hence, option (B) is the correct answer.


28. 22|xcosπx|dx is equal to 

(A) 8π 

(B) 4π.

(C) 2π                

(D)   1π                

Ans: Let I=22|xcosπx|dx

Here, it is a even function, therefore 

I=220|xcosπx|dx

I2=120xcosπxdx+3212xcosπxdx+232xcosπxdx

I2=[xπsinπx+1π2cosπx]012[xπsinπx+1π2cosπx]1232+[xπsinπx+1π2cosπx]320

I2=[12π1π2][32π12π]+[1π2+32π]

I2=12π1π2+32π+12π+1π2+32π

I2=82π

I=8π

Hence, option (A) is the correct answer.


Fill in the blanks in each of the  29 to 32.

29. sin6xcos8x dx= …………..

Ans: Let I=sin6xcos8x dx

I=sin6xcos6x.  cos2x dx

I=tan6x . sec2x dx

Put tanx=t

sec2x dx=dt

Therefore, 

I=t6  dt

I=t77+C

I=tan7x7+C

Hence, sin6xcos8x dx=tan7x7+C.


30. aaf(x)dx=0 if f is an ……………. function.

Ans: As we know that by the property of integration, aaf(x)dx=0 if f is an odd function.


31. 2a0f(x)dx=2 a0f(x)dx , if f(2ax)= ………..

Ans: As, we know that by the property of integration, we have  

2a0f(x)dx=2 a0f(x)dx , if f(2ax)=f(x) 

=0, if f(2ax)=f(x)

Hence, 2a0f(x)dx=2 a0f(x)dx , if f(2ax)=f(x).


32. π20sinnxsinnx + cosnxdx= …………

Ans: Let I=π20sinnxsinnx + cosnxdx …………….. (i)

I=π20sinn(π2x)sinn(π2x) + cosn(π2x)dx

I=π20cosnxcosnx + sinnxdx ……………… (ii)

Adding eq (i) and (ii), we get 

2I=π20dx

2I=[x]0π2

2I=π2

I=π4

Hence, π20sinnxsinnx + cosnxdx=π4.


Exercise 7.3

Short Answer (S.A)

Verify the following: 

1. 2x12x+3 dx=xlog|(2x3)2|+C

Ans: Let I= 2x12x+3 dx

I= 2x+3312x+3 dx

I= 2x+342x+3 dx

I= (142x+3) dx

I=  dx42x+3 dx

I= x42 (x+32) dx

 I= x2log(x+32)+C

 I= x2   log(2x+32)+C

  I= x2   [log(2x+3)log2]+C

  I= x2   log(2x+3)+2log2+C

I= x   log |(2x+3)2|+C

Thus, 2x12x+3 dx=xlog|(2x3)2|+C

Hence proved.


2. 2x+3x2+3x dx=log|x2+3x|+C

Ans: let I= 2x + 3x2+ 3x dx

Put x2+3x=t

(2x+3)dx=dt

I=1t dt 

I=log|t|+C

I=log|(x2+3x)|+C

Thus, 2x+3x2+3x dx=log|x2+3x|+C

Hence proved. 


Evaluate the Following: 

3. (x2+2)dxx+1

Ans: Let I=(x2+2)dxx+1 

I=(x1+3x+1)dx

I=xdxdx+3x+1dx

I=x22x+3 log|x+1|+C

Hence, (x2+2)dxx+1=x22x+3 log|x+1|+C.


4. e6logxe5logxe4logxe3logx dx

Ans: Let I=e6logxe5logxe4logxe3logx dx

I=elogx6elogx5elogx4elogx3 dx

I=x6x5x4x3 dx                                [elogx=x]

I=x5(x1)x3(x1) dx

I=x2 dx

I=x33+C

Hence, e6logxe5logxe4logxe3logx dx=x33+C.


5. (1+cosx)x+sinxdx

Ans: Let I=(1+cosx)x+sinxdx 

Put x+sinx=t

(1+cosx)dx=dt

I=1tdx

I=log|t|+C

I=log|(x+cosx)|+C


6. dx1+cosx

Ans: Let I=dx1+cosx

I=dx2cos2x2 

I=12dxcos2x2 

I=12sec2x2 dx

I=12×2tanx2 +C                [sec2x dx=tanx+C]

I=tanx2 +C

Hence, dx1+cosx=tanx2 +C.


7. tan2x. sec4x dx

Ans: Let I=tan2x . sec4x dx

I=tan2x . sec2x. sec2x dx

I=tan2x(1+tan2x )sec2x dx

Put tan   x=t

sec2  dx=dt

Therefore, 

I=t2(1+t2 )dt

I=(t2+t4)dt

I=t33+t55+C

I=tan3x3+tan5x5+C


8. sinx+cosx1+sin2xdx

Ans: Let I= sinx+cosx1+sin2xdx

 ⇒ I= sinx+cosxsin2x+cos2x+2.sinxcosxdx

I= sinx+cosx(sinx+cosx)2dx

I= (sinx+cosx)(sinx+cosx)dx

I= dx

I= x+C

Hence, sinx+cosx1+sin2xdx=x+C.


9. 1+sinx dx

Ans: Let I=1+sinx dx 

I=sin2x2+cos2x2+2sinx2cosx2 dx

I=(sinx2+cosx2)2 dx

I=(sinx2+cosx2) dx

I=sinx2 dx+ cosx2 dx

I=2cos   x2+   2   sin   x2+C

Hence, 1+sinx dx=2cos   x2+   2   sin   x2+C.


10. xx+1dx

Ans: Let I=   xx+1dx

Put x=z

12xdx=dz

dx=2x dz

dx=2z dz

Therefore,

I=   z2  +1×2  dz

I=   2z3  +1   dz

I=   2z3+11  +1   dz

I=   2z3+1  +1   dz   21  +1   dz

I=   2(  +1)(z2z+1)  +1   dz   21  +1   dz

I=   2(z2z+1)   dz   21  +1   dz

I=   2(z33z22+z)   2   log   |z+1|+C

I=   2(xx3x2+x)   2   log   |x+1|+C

I=   2(xx3x2+xlog   |x+1|)+C

Hence, xx+1dx= 2(xx3x2+xlog   |x+1|)+C .


11. a+xax dx

Ans: Let I=a + x a  x  dx

I=(a + x) (a + x) (a  x) (a + x)  dx

I=(a + x) 2 (a2x2)  dx

I=a + x  (a2x2)  dx

I=a   (a2x2)  dx+ x  (a2x2)  dx

I=a   sin1(xa)+ x  (a2x2)  dx

Put  a2x2=t2

2x dx=2t dt

x dx=t dt

Therefore, 

I=a   sin1(xa)+ t dt t2  

I=a   sin1(xa)t dt t 

I=a   sin1(xa) dt

I=a   sin1(xa)t+  C

I=a   sin1(xa)(a2x2) +  C


12. x121+ x34 dx

Ans: Let I=x121+ x34 dx

Put  x= z4

dx=4z3dz

Therefore, 

I=(z4)121+ (z4)34 ×4z3dz

I=4z2.z31+ z3 dz

Now, put 1+ z3=t

3z2dz=dt

z2dz=dt3

Therefore, we get

I=4(t1)t.dt3

I=43(11t )dt

I=43t43   log|t|+C1

I=43(1+z3)43   log|1+z3|+C1

I=43(1+x34)43   log|1+x34|+C1

I=43+43x3443   log|1+x34|+C1

I=43x3443   log|1+x34|+C


13. 1+ x2x4 dx 

Ans: Let I=1+ x2x4 dx

I=1+ x2x.  x3 dx

I=1+ x2x2 dxx3

I=(1x2+1) dxx3

Put (1x2+1)=t2

2x3dx=2t dt

dxx3= t dt

Therefore, 

I=t2 (t) dt

I=t. t dt

I=t2  dt

I=t33+C

I=13(1x2+1)32+C


14. dx169x2

Ans: Let I=dx169x2

I=dx9 (169x2)

I=13dx( 43)2x2

I=13 sin1(x43)+C

I=13 sin1(3x4)+C


15. dt3t 2t2

Ans: Let I=dt3t 2t2

I=dt2(t23t2)

I=12dt(t23t2)

I=12dt[t23t2+(34)2(34)2]

I=12dt[(t34)2(34)2]

I=12dt(34)2(t34)2

I=12 sin1(t3434)+C

I=12 sin1(4t33)+C


16. 3x1x2+9 dx

Ans: Let I=3x1x2+9   dx

I=3xx2+9   dxdxx2+9   

I=I1I2  ………………. (i),

Where,  I1=3xx2+9   dx  and    I2=dxx2+9   

Therefore, 

I1=3xx2+9   dx

Put x2+9=t2

2x dx=2t dt

x dx=t dt

 I1=3  dtt2

 I1=3  dtt

 I1=3 dt

   I1=3t+ C1  

   I1=3x2+9+ C1 ……….. (ii)

Also we have, 

   I2=dxx2+9    

   I2=dxx2+32    

   I2=log|x+x2+9|+C2  ………….. (iii)

Put the value of    I1 and    I2 in eq (i), we get 

I=3x2+9+ C1   log|x+x2+9|C2  

I=3x2+9   log|x+x2+9|+C


17. 52x+x2 dx

Ans: Let I= 52x+x2 dx

I= x22x+5 dx

I= x22x+1+4 dx

I= (x22x+1)+22 dx

I= (x1)2+22 dx

I=x12(x1)2+22+222 log|x1+(x1)2+22|+C  

I=(x1)252x+x2+2 log|x1+52x+x2|+C


18. xx41dx

Ans: Let I= xx41dx 

I= x(x2)21dx

Put x2=t.  

2x dx=dt

x dx=dt2

Therefore, 

I=12 dt(t)21

I=12.12×1log|t1t+1|+C

I=14   log|x21x2+1|+C


19. x21 x4 dx

Ans: Let I= x21 x4 dx

I= x22+x2212 (x2)2 dx

I= 12 +  x22 12+x22(1x2)(1 + x2) dx

I= 12(1 +  x2) 12(1x2)(1x2)(1 + x2) dx

I=12(1 +  x2)(1x2)(1 + x2) dx12(1  x2)(1x2)(1 + x2) dx

I=12dx(1x2) 12dx(1 + x2) 

I=12.12×1log|1+x1x |12 tan1x+C

I=14log|1+x1x |12 tan1x+C


20. 2ax x2 dx

Ans: Let I=2ax x2 dx

I=( x22ax) dx

I=( x22ax+a2a2) dx

I=[(x22ax+a2)a2] dx

I=[(xa)2a2] dx

I=a2(xa)2 dx

I=(xa)2a2(xa)2+a22 sin1(xaa)+C

I=(xa)22ax x2 +a22 sin1(xaa)+C


21. sin1x(1x2)32dx

Let I=sin1x(1x2)32 dx

I=sin1x(1  x2)1  x2 dx

Put  sin1x=t.   

11x2dx=dt

And sin1x=t  ⇒ x=sint

cost=1x2

Therefore, 

I=t.dt(1  sin2t) 

I=t.  dtcos2t 

I=t. sec2t dt

I=tsec2t dt(ddtt.sec2t dt)dt

I=  tanttantdt

I=  tant+log|cost|+C

I=sin1x   ×sintcost+log|1x2|+C

I=x sin1x1x2   +log|1x2|+C

Hence, sin1x(1x2)32 dx=x sin1x1x2   +log|1x2|+C.


22. (cos5x+cos4x)12cos3xdx 

Ans: Let I= (cos5x+cos4x)12cos3xdx

Applying [cosC+cosD=2cosC+D2. cosCD2 d cos2x=2cos2x1]

I= 2cos9x2 .cosx212 (2 cos23x21)dx

I= 2cos9x2 .cosx214 cos2 3x2+2dx

I= 2cos9x2 .cosx234 cos23x2dx

Now, multiply and divide by cos3x2, we get

I= 2cos9x2 .cosx2. cos3x2 3cos3x24 cos33x2dx

I= 2cos9x2 .cosx2. cos3x2 4 cos33x23cos3x2dx

I= 2cos9x2 .cosx2. cos3x2 cos3.3x2dx

I= 2cos9x2 .cosx2. cos3x2 cos9x2dx

I= 2cosx2. cos3x2dx

I= [cos(3x2+x2)+cos(3x2x2)]dx

I= (cos2x+cosx)dx

I= cos2xdxcosxdx

I= (sin2x2)sinx +C

Hence, (cos5x+cos4x)12cos3xdx=(sin2x2)sinx +C.


23. sin6x + cos6x sin2x .cos2xdx

Ans: Let I= sin6x + cos6x sin2x .cos2xdx

I= (sin2x)3 + (cos2x)3 sin2x .cos2xdx

I= (sin2x+ cos2x  ) (sin4xsin2x .cos2x+cos4x) sin2x .cos2xdx

I=  (sin4xsin2x .cos2x+cos4x) sin2x .cos2xdx

I=  sin4xsin2x .cos2xdx sin2x .cos2xsin2x .cos2xdx+ cos4xsin2x .cos2x dx

I=  sin2xcos2xdxdx+ cos2xsin2x  dx

I= tan2x dxdx+cot2x dx

I= (sec2x1 ) dxdx+(cosec2x1)  dx

I= (sec2x dx+cosec2x dx3dx

I= tanxcotx3x+C

Hence, sin6x + cos6x sin2x .cos2xdx= tanxcotx3x+C.


24. xa3x3dx

Ans: Let I=xa3x3dx

I=x12(a32)2(x32)2dx

Here, Put   x32=t

32 x12 dx=dt

 x12 dx=23dt

Therefore, 

I=23dt(a32)2(t)2

I=23 sin1(ta32)+C

I=23 sin1(  x32a32)+C

I=23 sin1(xa)32+C

Hence, xa3x3dx=23 sin1(xa)32+C.


25. cos xcos 2x1cosx dx

Ans: Let I=cos xcos 2x1cosx dx

I=2 sin 3x2.  sin x22 sin2x2 dx

I= sin 3x2  sin x2 dx

I= 3 sin x24 sin3 x2  sin x2 dx

I=3dx4 sin2 x2 dx

I=3x4 1+cos x2 dx

I=3x2 (1cos x) dx

I=3x2(xsin x)+C

I=3x2x+2 sin x+C

I=x+2 sin x+C

Hence, cos xcos 2x1cosx dx=x+2 sin x+C.


26. dxxx41

Ans: Let I=dxxx41

Multiply and divide by x3, we get

I=x3dxx4x41

Now, put  x41= t2

4x3dx=2t dt

x3dx=t2 dt

Therefore, 

I=12t dt(1+t2)t

I=12 dt(1+t2)

I=12 tan1t+C

I=12 tan1(x41)+C

Hence, dxxx41=12 tan1(x41)+C.


Evaluate the Following As Limit of Sums: 

27. 20(x2+3) dx

Ans: Let I=20(x2+3) dx 

Here, we have a=0, b=2 and h=ban=2n

nh=2 and f(x)=(x2+3)

Now, 20(x2+3) dx=limh0h[f(0)+f(0+h)++f{0+(n1)h}]

=limh0h[3+h2+3+22h2+3++(n1)2h2+3] 

=limh0h[3n+h2{12+22+32++(n1)2}] 

=limh0h[3n+h2× n(n1)(2n1)6] 

=limh0[3nh+h3× n(n1)(2n1)6] 

=limh0[3nh+ nh(nhh)(2nhh)6] 

=limh0[3.2+ 2(2h)(2.2h)6] 

=6+4×46 

=6+166 

=6+83 

=263 

Hence, I=20(x2+3) dx=263


28. 20ex dx

Ans: let I=20ex dx

Here, we have a=0, b=2 and h=ban=2n

nh=2 and f(x)=ex

Now, 20ex dx=limh0h[f(0)+f(0+h)++f{0+(n1)h}]

=limh0h[1+eh+e2h+e3h++e(n1)h] 

=limh0h[1.  (eh)n1eh1] 

=limh0h[enh1eh1] 

=limh0h[e21eh1] 

=e2 limh0heh1 limh0heh1 

=e21 

Hence, I=20ex dx=e21 .


29. π20tanx dx1+ m2 tan 2x

Ans: Let I= π20tanx dx1+ m2 tan 2x

I= π20sinxcos x dx1+ m2sin 2xcos2x

I= π20sinxcos x dxcos2x+m2sin 2xcos2x

I= π20sin x. cosx dxcos2x+m2sin 2x

I= π20sin x. cosx dx1sin 2x(1m2)

Now, put sin2x=t

2.sinx.cosx dx=dt

Here, at x=0t=0 and at x=π2t=1

Therefore, 

I=1210dt1t(1m2)

I=12 [log|1t(1m2)|×11m2]01 

I=12 ×11m2[log|1t(1m2)|]01

I=12(1m2)[log|11(1m2)|log|10(1m2)|] 

I=12(1m2)[log|11+m2|log|1|] 

I=12(1m2)[log|m2|]

I=12(1m2)2×log|m|

I=1(1m2)log|m|

Hence, π20tanx dx1+ m2 tan 2x=1(1m2)log|m| .


30. 21dx(x1)(2x)

Ans: Let I=21dx(x1)(2x)

I=21dx2xx22+x

I=21dxx2+3x2

I=21dx(x23x+2)

I=21dx[x23x+(32)2(32)2+2]

I=21dx[{x23x+(32)2}{942}]

I=21dx[(x32)2(12)2]

I=21dx[(12)2(x32)2]

I=[sin1(x3212)]12

I=[sin1(2x3)]12

I=[sin1(43)sin1(23)]

I=[sin1(1)sin1(1)]

I=[π2(π2)]

I=[π2+π2]

I=π

Hence, 21dx(x1)(2x)=π


Evaluate the Following: 

31. 10dxex+ex

Ans: Let I=10dxex+ex

I=10dxex+1ex

I=10ex.dx(ex)2+1

Now, put ex=t

exdx=dt

Now, at x=0 t=1 and at x=1 t=e

Therefore, 

I=e1dtt2+1

I=[tan1t]1e

I=(tan1etan11)

I=tan1e π4

Hence, 10dxex+ex=tan1e π4


32. 10x dx1+x2

Ans: Let I=10x dx1+x2

Now, put 1+x2=t2

2x dx=2t dt

x dx=t dt

here, at x=0 t=1 and at x=1 t=2

Therefore, 

I=21t dtt2

I=21t dtt

I=21dt

I=[t]12

I=(21)

Hence, 10x dx1+x2=(21)


33. π0xsinx cos2x dx

Ans: Let I=π0xsinx cos2x dx ………. (i)

I=π0(πx)sin(πx) cos2(πx) dx

I=π0(πx)sinx cos2x dx …………(ii)

Adding eq(i) and (ii), we get 

2I=ππ0sinx cos2x dx

Now, put cosx=t

sinx dx=dt

sinx dx= dt

here, at x=0 t=1 and at x=π t=1

Therefore, 

2I=π11t2dt

2I=π[t33]11

2I=13π[t3]11

2I=13π (11)

⇒ 2I=2π3

I=π3

Hence, π0xsinx cos2x dx=π3 .


34. 120dx(1+x2)1x2

Ans: Let I= 120dx(1+x2)1x2

Put x=sinθ

dx=cosθ dθ

Here, at x=0 θ=0 and at x=12 θ=π6

Therefore, 

I= π60cos θ dθ(1+sin2θ)1sin2θ

I= π60cos θ dθ(1+sin2θ)cos θ

I= π60 dθ(1+sin2θ)

Divide by cos2θ in numerator and denominator 

I= π601cos2θ dθ(1cos2θ+sin2θcos2θ)

I= π60sec2θ dθ(sec2θ+tan2θ)

I= π60sec2θ dθ(1+tan2θ+tan2θ)

I= π60sec2θ dθ(1+2tan2θ)

Now, put tanθ=t

sec2θ dθ=dt

Here, at θ=0t=0 and at θ=π6t=13

Therefore, 

I= 130dt(1+2t2)

I=12 130dt(12+t2)

I=12 130dt[(12)2+t2]

I=12 [112 tan1(t12)]013

I=12 [ tan1(2 ×t)]013

I= 12[tan1(2 ×13)tan1(0)]

I= 12tan1( 23)


Long Answer (L.A)

35. x2 dxx4x212

Ans: Let I= x2 dxx4x212

I= x2 dxx44x2+3x212

I= x2 dxx2(x24)+3(x24)

I= x2 dx(x24)(x2+3)

I= x24+ 4 dx(x24)(x2+3)

I= (x24) dx(x24)(x2+3)+4dx(x24)(x2+3) 

I=  dx(x2+3)+4[(x24)(x2+3)}dx(7)(x24)(x2+3) 

I=  dx(x2+3)47(x24)dx(x24)(x2+3)+47(x2+3)dx(x24)(x2+3) 

I=  dx(x2+3)47dx(x2+3)+47dx(x24) 

I= 37dx(x2+3)+47dx(x24) 

I= 37dx(x2+(3)2)+47dx(x222) 

I= 37[13 tan1x3]+47.12×2log|x2x+2|+C 

I= 37 tan1x3+17log|x2x+2|+C 

Hence, x2 dxx4x212=37 tan1x3+17log|x2x+2|+C.


36. x2 dx(x2+a2) (x2+b2) 

Ans: Let I= x2+a2a2 dx(x2+a2) (x2+b2) 

I= (x2+a2) dx(x2+a2) (x2+b2)  a2 dx(x2+a2) (x2+b2) 

I=  dx (x2+b2)  a2[(x2+a2)(x2+b2)] dx(a2b2)(x2+a2) (x2+b2) 

I=  dx (x2+b2) a2(a2b2)[(x2+a2)(x2+b2)] dx(x2+a2) (x2+b2) 

I=  dx (x2+b2) a2(a2b2)[(x2+a2) dx(x2+a2) (x2+b2) (x2+b2) dx(x2+a2) (x2+b2) ]

I=  dx (x2+b2) a2(a2b2)[ dx (x2+b2)  dx(x2+a2)  ]

I=b2(a2b2)  dx (x2+b2) +a2(a2b2) dx(x2+a2)  

I=b2(b2a2)1btan1xba2(b2a2)1a tan1xa+C

I=b(b2a2)tan1xba(b2a2) tan1xa+C


37. π0x1+sinxdx

Ans: Let I=π0x1+sinxdx  ……….(i)

I=π0(πx)1+sin(πx)dx

I=π0(πx)1+sinxdx ………….(ii)

Now, adding eq(i) and (ii), we get

2I=π0 π 1+sinxdx 

2I= π π0(1sinx)(1+sinx)(1sinx)dx 

2I= π π0(1sinx)(1sin2x)dx 

2I= π π0(1sinx)cos2xdx 

2I= π π0(1cos2xsinxcos2x)dx 

2I= π π0(sec2xsecx.tanx)dx 

2I= π [tanxsecx]0π 

2I= π [tanπsecπ (tan0sec0 )]0π 

2I= π [0(1)(01)] 

2I= π [1+1] 

2I=2 π  

I= π  

Hence, π0x1+sinxdx=π


38. 2x1(x1)(x+2)(x3)dx

Ans: Let I= 2x1(x1)(x+2)(x3)dx

Now, 2x1(x1)(x+2)(x3)=A(x1)+B(x+2)+C(x3)

2x1=A(x+2)(x3)+B(x1)(x3)+C(x1)(x+2)

Here, put x=1 in above equation, we get

1=A(1+2)(13)+0+0

1=6 A

A=16 

Now, put x=2

2(2)1=0+B(21)(23)+0

5=B(3)(5)

B=13

Now, put x=3, we get

2(3)1=0+0+C(31)(3+2)

5=C(2)(5)

C=12

Therefore, 

2x1(x1)(x+2)(x3)dx=dx6(x1)dx3(x+2)+dx2(x3) 

=16log|(x1)|13log|(x+2)|+12log|(x3)|+C 

=log(x1)16log(x+2)13+log(x3)12+C 

=log|(x3)(x1)16 (x+2)13|+C 

Hence, 2x1(x1)(x+2)(x3)dx=log|(x3)(x1)16 (x+2)13|+C


39. etan1x(1+x+x21+x2)dx

Ans: Let I=etan1x(1+x+x21+x2)dx

I=etan1x(1+x21+x2+x1+x2)dx

I=etan1x(1+x1+x2)dx

I=etan1xdx+ xetan1x1+x2dx

I=I1+I2 …...(i), where, I1=etan1xdx  &  I2= xetan1x1+x2dx

Now, I2= xetan1x1+x2dx

Put tan1x=t

11+x2dx=dt

Therefore, 

I2= et.tant dt

I2= et.tant dt

I2=tantet dt(ddxtant.et dt)dt

I2=tant. etsec2t.  etdt+C

I2=x. etan1x(1+tan2t).  etan1x11+x2dx+C

I2=x. etan1x(1+x2).  etan1x11+x2dx+C

I2=x. etan1xetan1xdx+C 

Put values of I1 and I2, in equation (i), 

I=etan1xdx+x. etan1xetan1xdx+C

I=x. etan1x+C

Hence, value of etan1x(1+x+x21+x2)dx=x. etan1x+C.


40. sin1xa+xdx

Ans: Let I= sin1xa+x  dx

Put x=a tan2θ

dx=a. 2tanθ.sec2θ dθ

Therefore, 

I= sin1a tan2θa+a tan2θ  ×  2atanθ.sec2θ dθ

I= sin1a tan2θa(1+ tan2θ)  ×  2atanθ.sec2θ dθ

I= sin1 tan2θ(1+ tan2θ)  ×  2atanθ.sec2θ dθ

I= sin1 tan2θsec2θ  ×  2atanθ.sec2θ dθ

I= sin1sin2θ  ×  2atanθ.sec2θ dθ

I= sin1(sinθ) ×  2atanθ.sec2θ dθ

I= 2aθ tanθ.sec2θ dθ 

I= 2a [θ tanθ.sec2θ dθ(ddθθ.  tanθ.sec2θ dθ)dθ] 

Put tanθ=t

sec2θ dθ=dt

I= 2a [θ t dt(  t dt)dθ] 

I= 2a [θ. t22t22dθ] 

I= 2a [θ. tan2θ2tan2θ2dθ] 

I= a [θ(tan2θ)   tan2θ dθ] 

I= aθ(tan2θ) a(sec2θ1) dθ

I= aθ(tan2θ)atanθ+ aθ+C 

I= a[xa(tan1xa)xa+ tan1xa]+C 

Hence, sin1xa+x  dx=a[xa(tan1xa)xa+ tan1xa]+C .


41. π2π31+cosx(1cosx)52 dx

Ans: Let I=π2π31+cosx(1cosx)52 dx

I=π2π32cos2x2(2 sin2x2)52 dx

I=2252π2π3cos   x2sin5x2 dx

Put sinx2=t

12cosx2 dx=dt

cosx2 dx=2dt

Here, at x=π3 t=12 and at x=π2 t=12

Therefore, 

I=23212122 dtt5 

I=242×21212 t5dt 

I=12 [t44]1212

I=18 [t4]1212

I=18 [(12)4(12)4]

I=18 [(2)4(2)4]

I=18 [416]

I=18 ×12

I=32 

Hence, π2π31+cosx(1cosx)52 dx=32 .


42. e3xcos3x dx

Ans: Let I= e3xcos3x dx

I= e3x(cos3x+3cosx4) dx

I=14(e3xcos3x+3e3xcosx) dx

I=14e3xcos3x dx+34 e3xcosx dx

I=14   I1+34   I2 ………… eq (i)

Where, I1=e3xcos3x dx and I2=3e3xcosx dx

Now, I1=e3xcos3x dx

I1=cos3xe3x dx(ddxcos3x .e3x dx)dx

I1=e3x3cos3x(3   sin3x .e3x3)dx

I1=e3x3cos3x(e3x   sin3x .)dx

I1=e3x3cos3x[sin3xe3xdx(ddxsin3x .e3x dx)dx]

I1=e3x3cos3x[e3x3sin3x(3cos3x .e3x3)dx]

I1=e3x3cos3x[e3x3sin3x+(e3xcos3x )dx]

I1=e3x3cos3x[e3x3sin3x+I1]

I1=e3x3cos3x+e3x3sin3xI1

2I1=e3x3cos3x+e3x3sin3x

2I1=13e3x(cos3x+sin3x)+C1

I1=16e3x(sin3xcos3x)+C1

Now, I2=e3xcosx dx

I2=e3xcosx dx

I2=cosxe3x dx(ddxcosx.e3x dx )dx

I2=e3x3cosx(sinx.e3x3 )dx

I2=e3x3cosx13(e3x.sinx )dx

I2=e3x3cosx13[sinxe3xdx(ddxsinx.e3xdx)dx]

I2=e3x3cosx13[e3x3sinx(e3x3cosx)dx]

I2=e3x3cosx13[e3x3sinx+13(e3xcosx)dx]

I2=e3x3cosx+19e3xsinx19(e3xcosx)dx

I2=e3x3cosx+19e3xsinx19I2

I2+I29=e3x3cosx+19e3xsinx

10I29=3e3x.cosx+e3xsinx9+C2

10I2=e3x(sinx3cosx)+C2

I2=110e3x(sinx3cosx)+C2

Now, put the value of I1 and I2 in eq(i), we get

I=124e3x(sin3xcos3x)+340e3x(sinx3cosx)+C   


43. tanx dx

Ans: Let I=tanx dx

Put tanx= t2

sec2x dx=2t dt

(1+tan2x) dx=2t dt

 dx=2t(1+tan2x)  dt

 dx=2t(1+t4)  dt

Therefore, 

I=t2 . 2t(1+t4)  dt

I= 2t2(1+t4)  dt

I= (1+t2)(1t2)(1+t4)  dt

I= (1+t2)(1+t4)  dt (1t2)(1+t4)  dt

I= (1t2 +1)(1t2 +t2)  dt (1t21) (1t2 +t2) dt

I= (1t2 +1)(t 1t)2+2  dt (1t21) (t+ 1t)22 dt

Put u=(t 1t) du= (1t2+1)dt

And v=(t+1t) du= (1t21)dt

Therefore,

I=  du(u)2+2   dv (v)22 

I=  du(u)2+(2)2 + dv (v)2(2)2 

I=12 tan1u2+122log|v2v+2|+C 

I=12 tan1(t 1t)2+122log|(t + 1t)2(t + 1t)+2|+C

I=12 tan1(tanx 1tanx)2+122log|(tanx + 1tanx)2(tanx + 1tanx)+2|+C

I=12 tan1(tanx 1)2tanx+122log|tanx2tanx +1tanx + 2tanx +1|+C


44. π20dx(a2cos2x+ b2sin2x)2

Ans: Let I= π20dx(a2cos2x+ b2sin2x)2

Divide numerator and denominator by cos4x

I= π20dxcos4x(a2cos2x+ b2sin2x)2cos4x

I= π20sec4x dx(a2+ b2tan2x)2

I= π20( 1+ tan2x) sec2x dx(a2+ b2tan2x)2

Put btanx=atant

 b sec2x dx=a sec2t dt

  sec2x dx=ab sec2t dt

When x0, t0 and When xπ2,tant0 , tπ2

Therefore, 

I= π20(1+ (abtant)2)a4sec4tab sec2t dt

I=1a3b3π20(b2+a2tan2t)cos2t dt

I=1a3b3π20(b2cos2t+a2sin2t) dt

I=1a3b3π20(b2 1+cos2t2+a2 1cos2t2) dt

I=12a3b3π20[(a2+b2)+( b2a2)cos2t]dt

I=12a3b3[(a2+b2)x12( b2a2)sin2t ]0π2

I=12a3b3[(a2+b2)π212( b2a2)sinπ]

I=12a3b3[(a2+b2)π2]

I=π(a2+b2)4a3b3 

Hence, π20dx(a2cos2x+ b2sin2x)2=π(a2+b2)4a3b3 .


45. 10xlog(1+2x) dx

Ans: Let I=10xlog(1+2x) dx

I=[log(1+2x)10x dx]0110(ddxlog(1+2x).10x dx)dx

I=[x22log(1+2x)]0110(21 + 2x.x22)dx

I=(12log30)10(x21 + 2x)dx

I=(12log30)10(x2x21 + 2x)dx

I=(12log3)1210xdx+1210x1+2xdx

I=12log312(120)+14101 + 2x11+2xdx

I=12log314+14101+ 2x1+2xdx1410dx1+2x

I=12log314+1410dx14[12log(1+2x)]01

I=12log314+14(x)0118[log(3)0]

I=12log314+1418[log(3)]

I=12log318log3

I=38log3

Hence, 10xlog(1+2x) dx=38log3.


46. π0xlogsinx dx

Ans: Let I=π0xlogsinx dx ……………….. (i)

I=π0(πx)logsin(πx) dx

I=π0(πx)logsinx dx …………..(ii)

Adding eq (i) and (ii), we get

2I=ππ0logsinx dx

2I=2ππ20logsinx dx

I=ππ20logsinx dx ………………..(iii)

I=ππ20logsin(π2x) dx

I=ππ20logcosx dx ……………(iv)

Adding eq (iii) and (iv), we get

2I=ππ20(logsinx+logcosx) dx

2I=ππ20(logsinx.cosx)dx

2I=ππ20(log22sinx.cosx)dx

2I=ππ20(logsin2x2)dx

2I=ππ20(logsin2xlog2)dx

2I=ππ20logsin2x dxππ20log2dx

2I=ππ20logsin2x dxπlog2 ×π2

Put 2x=t dx=dt2

Here, at x=0t=0 and at x=π2t=π

2I=π2π0logsint dtπ22 log2 

2I=π2×2π20logsint dtπ22 log2

2I=ππ20logsint dtπ22 log2

2I=Iπ22 log2                           [from eq (iii)]

I=π22 log2

Hence, π0xlogsinx dx=π22 log2.


47. π4π4log(sinx+cosx)dx 

Ans: Let I=π4π4log(sinx+cosx)dx ……….. (i)

I=π4π4log[sin(π4π4x)+cos(π4π4x)]dx

I=π4π4log[sin(x)+cos(x)]dx

I=π4π4log(sinx+cosx)dx 

I=π4π4log(cosxsinx)dx …………..(ii)

Adding eq (i) and (ii),we get

⇒ 2I=π4π4log(sinx+cosx)dx+π4π4log(cosxsinx)dx 

⇒ 2I=π4π4log[(sinx+cosx).(cosxsinx)]dx 

⇒ 2I=π4π4log[cos2xsin2x]dx

⇒ 2I=π4π4logcos2xdx

⇒ 2I=2π40logcos2xdx

I=π40logcos2xdx  ……………(iii)

Now, put 2x=tdx=dt2

Here, at x=0t=0 and at x=π4t=π2

Therefore, 

I=12π20logcost dt ………… (iv)

I=12π20logcos(π2t) dt

I=12π20logsint dt …………..(v)

Adding eq (iv) and (v), we get

2I=12π20logcost dt+12π20logsint dt

2I=12π20log   sintcost dt

4I=π20log22sintcost dt

4I=π20logsin2t2 dt

4I=π20(logsin2tlog2) dt

4I=π20(logsin(π22t)) dtπ20log2 dt

4I=π20(logcos2t) dtπ2log2

4I=2π40(logcos2t) dtπ2log2

4I=2Iπ2log2  

2I=π2log2

I=π4log2

Hence, π4π4log(sinx+cosx)dx=π4log2.


Objective Type Questions 

Choose the correct option from given four options in each of the Exercises from 48 to 63.

48. cos2xcos2θ cosxcosθdx is equal to

(A) 2(sinx+xcosθ)+C                    

(B) 2(sinxxcosθ)+C       

(C)  2(sinx+2xcosθ)+C                   

(D) 2(sinx2xcosθ)+C

Ans: Let I= cos2xcos2θ cosxcosθdx  

I= 2 cos2x1(2 cos2θ1)cosxcosθdx

I= 2 cos2x12 cos2θ+1cosxcosθdx

I= 2 cos2x2 cos2θcosxcosθdx

I= 2( cos2x cos2θ)(cosxcosθ)dx

I= 2( cos2x cos2θ)(cosxcosθ)dx

I= 2(cosx+ cosθ)dx

I= 2(sinx+xcosθ)+C

Hence, option (A) is correct answer.


49. dxsin(xa)sin(xb) is equal to 

(A) sin(ba)log|sin(xb)sin(xa)|+C    

(B) cosec(ba)log|sin(xa)sin(xb)|+C

(C) cosec(ba)log|sin(xb)sin(xa)|+C            

(D) sin(ba)log|sin(xb)sin(xa)|+C

Ans: Let I= dxsin(xa)sin(xb) 

I=1sin(ba)sin(ba) dxsin(xa)sin(xb) 

I=1sin(ba)sin(xa x + b) sin(xa)sin(xb)dx 

I=1sin(ba)sin{(xa)( x b)} sin(xa)sin(xb) dx

I=1sin(ba)sin(xa).cos(xb) sin(xb).cos(xa)  sin(xa)sin(xb) dx

I=1sin(ba)cos(xb)sin(xb)dx1sin(ba)cos(xa)sin(xa)dx

I=1sin(ba)cot(xb)dx1sin(ba)cot(xa)dx

I=1sin(ba)log|sin(xb)|1sin(ba)log|sin(xa)|+C

I=1sin(ba)[log|sin(xb)|log|sin(xa)|]+C

I=1sin(ba)log|(xb)(xa)|+C

I=cosec(ba)log|(xb)(xa)|+C

Hence, option (C) is correct answer.


50. tan1x dx is equal to 

(A) (x+1)tan1xx+C       

(B)  xtan1xx+C      

(C) xxtan1x+C                 

(D) x(x+1)tan1x+C

Ans: Let I=tan1x dx

I=1. tan1x dx

I= tan1x 1 dx(ddxtan1x .1 dx)dx

I= xtan1x (11+x.12x .x)dx

I= xtan1x 12(x1+x)dx

Now, put  x=t2

dx=2t dt

Therefore, 

I= xtan1x 12t1+t2. 2t dt

I= xtan1x t21+t2 dt

I= xtan1x 1+t211+t2 dt

I= xtan1x (111+t2) dt

I= xtan1x 1 dt+(11+t2) dt

I= xtan1x t+tan1t+C

I= xtan1x x+tan1x+C

I=(x+1)tan1x x+C

Hence, option (A) is correct answer.


51. ex(1x 1+x2)2dx is equal to

(A)  ex1+x2+C                                           

(B) ex1+x2+C            

(C) ex(1+x2)2+C                                      

(D) ex(1+x2)2+C                                                                                              

Ans: Let I=ex(1x 1+x2)2dx

I=ex[1+x22x (1+x2)2]dx

I=ex[1+x2 (1+x2)22x (1+x2)2]dx

I=ex[1 (1+x2)2x (1+x2)2]dx

I=ex (1+x2)+C

Hence, option (A) is correct answer.


52. x9(4x2+1)6dx is equal to

(A) 15x(4+1x2)5+C                   

(B) 15(4+1x2)5+C        

(C) 110x(1+4)5+C                    

(D) 110(4+1x2)5+C                                  

Ans: Let I=x9(4x2+1)6dx

I=x9[x2(4+1x2)]6dx

I=x9x12(4+1x2)6dx

I=dxx3(4+1x2)6

Now, put 4+1x2=t

2x3dx=dt

dxx3=dt2

Therefore, 

I=12dt(t)6

I=12t6 dt

I=12 × t55+C

I=110(4+1x2)5+C

Hence, option (D) is correct answer.


53. dx(x+2)(x2+1)=alog|1+x2|+b tan1x+15log|x+2|+C,   then

(A). a=110 , b=25                    

(B) a=110 , b=25

(C) a=110 , b=25                          

(D) a=110 , b=25

Ans: Let I=dx(x+2)(x2+1) 

Now, 1(x+2)(x2+1)=A(x+2)+Bx+C(x2+1)

1(x+2)(x2+1)=A(x2+1)+(Bx+C) (x+2)(x2+1)

1=A(x2+1)+(Bx+C) (x+2)

1=Ax2+A+Bx2+2Bx+Cx+2C

1=x2(A+B)+x(2B+C)+(A+2C)

On comparing both sides, we get

A+B=0,  2B+C=0 and A+2C=1

From above equations, we get

A=15 , B=15  and C=25

Therefore, 1(x+2)(x2+1)=15(x+2)+x5+25(x2+1)

dx(x+2)(x2+1)=dx5(x+2)+x5+25(x2+1)dx

dx(x+2)(x2+1)=dx5(x+2)15x(x2+1)dx+251(x2+1)dx

=15log|x+5|110log|1+x2|+25tan1x+C 

Here, a=110 and b=25

Hence, option (C) is correct answer.


54. x3x+1dx is equal to 

(A) x+x22+x33log|1x|+C    

(B) x+x22x33log|1x|+C

(C) xx22x33log|1+x|+C

(D) xx22+x33log|1+x|+C

Ans: Let I=x3x+1dx

I=x3+11x+1dx

I=x3+1x+1dx1x+1dx

I=(x2x+1)dx1x+1dx

I=x33x22+xlog|x+1|+C

Hence, option (D) is correct answer.


55. x+sinx1+cosx dx is equal to 

(A) log|1+cosx|+C                

(B) log|x+sinx|+C

(C) xtanx2+C                           

(D) xtanx2+C                           

Ans: Let I=x+sinx1+cosx dx

I=x1+cosx dx+sinx1+cosx dx

I=I1+I2   ………(i), where, I1=x1+cosx dx & I2=sinx1+cosx dx

Now, I1=x1+cosx dx

I1=x2 cos2x2 dx

I1=12x sec2x2 dx

I1=12[x sec2x2 dx (ddxx. sec2x2 dx  )dx]

I1=12[2xtanx2 2tanx2 dx]+C

I1=xtanx2 tanx2 dx+C …………. (ii)

And also we have,

I2=sinx1+cosx dx

I2=2sinx2 .cosx22 cos2x2 dx

I2=sinx2 cosx2 dx

I2=tanx2 dx …………..(iii)

Now, put the value of I1 and I2 in eq (i), 

I=xtanx2 tanx2 dx+C+tanx2 dx 

I=xtanx2+C

Hence, option (D) is correct answer.


56. x3 dx1+ x2=a(1+x2)32+b1+x2+C, then 

(A) a=13 , b=1                          

(B) a=13 , b=1               

(C) a=13 , b=1                    

(D) a=13 , b=1                                                         

Ans: Let I=x3 dx1+ x2

I=x2.x dx1+ x2

Put 1+ x2=t2

2x dx=2t dt

x dx=t dt

Therefore, 

I=(t21)t dtt

I=(t21)dt

I=t33t+C

I=(1+x2)331+x2+C

I=(1+x2)3231+x2+C

Here, a=13 , b=1

Hence, option (D) is correct answer.


57 π4 π4 dx1+cos2x is equal to

(A) 1

(B) 2

(C) 3

(D)  4

Ans: Let I=π4 π4 dx1+cos2x

I=π4 π4 dx2 cos2x

I=12π4 π4 sec2xdx

I=12 [tanx]π4 π4 

I=12 [tanπ4 tan(π4) ]

I=12 [1+1]

I=1

Hence, option (A) is correct answer.


58. π201sin2x dx is equal to

(A) 22            

(B) 2(2+1)        

(C) 2            

(D) 2(21)        

Ans: Let I= π201sin2x dx

I= π20sin2x+cos2x2sinx.cosx  dx

I= π20(sinxcosx)2  dx

I= π20 |sinxcosx| dx

Since, (sinxcosx)0, x[0, π4] and (sinxcosx)0, x [π4,π2]

Therefore, 

I= π40 (sinxcosx)dx+π2π4 (sinxcosx)dx

I= [cosxsinx]0π4+[cosxsinx]π4π2

I= [cosx+sinx]0π4[cosx+sinx]π4π2

I= [cosπ4+sinπ410][cosπ2+sinπ2cosπ4sinπ4]

I= [12+121][11212]

I= [21][12]

I= 211+2

I= 222

I= 2(21)

Hence, option (D) is correct answer.


59. π20cosx esinxdx is equal to …………..

Ans: Let I=π20cosx esinx dx

Put sinx=t

cosxdx=dt

Now, at x=0t=0 and at x=π2t=1

Therefore, 

I=10et dx 

I=[et]01

I=e1e0

I=e1

Hence, π20cosx esinx dx=e1.


60. x+3 (x+4)2 exdx = ………

Ans: Let I=x+3 (x+4)2 exdx

I=ex[x+41 (x+4)2] dx

I=ex[x+4 (x+4)21 (x+4)2] dx

I=ex[1 x+41 (x+4)2] dx

I=ex x+4+C

Hence, x+3 (x+4)2 exdx=ex x+4+C.


Fill in the blanks in each of the following Exercise 60 to 63. 

61. a011+4x2dx=π8, then a= ………..

Ans: here, we have a011+4x2dx=π8

14a0114+x2dx=π8

a01(12)2+x2dx=π2

[112 tan1(x12)]0a=π2

[2 tan1(2x)]0a=π2

tan1(2a)tan1(0)=π4

tan1(2a)=π4

2a=tanπ4

2a=1

a=12

Hence, value of a is 12 .

Therefore, a011+4x2dx=π8, then a=12 .


62. sinx3+4 cos2xdx=

Ans: Let I=sinx3+4 cos2xdx

Put cosx=t 

sinx dx=dt

sinx dx=dt

Therefore, 

I=dt3+4 t2

I=14dt34+ t2

I=14dt(32)2+ t2

I=14 . 132tan1(t32)+C

I= 123tan1(2t3)+C

I= 123tan1(2cosx3)+C

Hence, sinx3+4 cos2xdx= 123tan1(2cosx3)+C.


63. The value of ππsin3x.cos2x dx is …………..

Ans: Let I=ππsin3x.cos2x dx …………. (i)

I=ππsin3(ππx).cos2(ππx) dx

I=ππsin3(x).cos2(x) dx

I=ππsin3x.cos2x dx …………….(ii)

Now, adding eq (i) and (ii), we get

2I=0

I=0

Hence, the value of ππsin3x.cos2x dx is 0.


Importance of Integrals

Students will get a brief knowledge about the concepts that are mentioned in the PDF. The Exemplar PDF of Chapter 7 - Integrals is designed as per the latest syllabus pattern advised by the Central Board of Secondary Education. The PDF is structured by the expert faculty of Vedantu, specifically structured to enhance the learning and understanding capacity of the students. The students following the Exemplar with full dedication will surely achieve good results. 


Class 12 is the most important academic year for students. All the Chapters in all the subjects are very important, students do have pressure to perform well in the Exams. Chapter 7 Integrals is such an important Chapter that focuses on the concepts used in other Chapters as well. This Chapter increases the thinking process of the students, questions asked from this Chapter are not difficult so this Chapter can be quite scoring for all students. 


Students neglecting the concepts of this Chapter face trouble as the concepts present in this Chapter also help to solve other Chapter questions as well. NCERT Exemplar of Chapter 7 consists of a textbook solution for Class 12. There are many important concepts but the most important concepts are mentioned below. This Chapter basically focuses on the questions asked on topics like -

  1. Integrations

  2. Properties of Integrals

  3. Geometrical representation

  4. Comparison between integration and differentiation 

  5. Definite Integral - limit of the sum

  6. Different

  7.  methods of integration

  8. Properties of definite Integral and some other important question


Those students who want to have a good grip on the concepts must focus on the structured solutions mentioned in the Exemplar. Students can register themselves on Vedantu’s website for other important topics. Vedantu also offers online study support to the students, the online present material consists of additional questions, notes, sample papers, Exemplars. All the study material is designed and approved by the community of faculty that consists of a group of teachers that focuses on the quality of the content present from the Vedantu’s end.

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FAQs on NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

1. From where do students can get the Exemplar for Chapter 7 Integral for Class 12?

The NCERT Exemplar for Chapter 7 Integral of Class 12 is available on Vedantu.com. Students reach out to the website and can get a free PDF for Chapter 7. The PDF of Chapter 7 - Integrals consists of important topics and additional questions that will help students to learn and practice to score more marks in their Examinations. The PDF is available at Vedantu’s official website for free. 

2. Is Chapter 7 important for Class 12 students?

Class 12 is the most important Class for students as well as the toughest Class. Students do have mental pressure to perform well in their boards. All the chapters of every subject are important as Chapter 7 Integrals. This Chapter increases the thinking process of the students, questions asked from this Chapter are not really tough though this Chapter can be a score booster for all students. Students neglecting the concepts of this Chapter go through problems as the concepts present in the respective Chapter help to solve other Chapter questions as well. 

3. What are the important concepts for the Chapter 7 Integrals?

Chapter 7 Integrals do have many important concepts and the most important topics are -Integrations, Properties of Integrals, Geometrical representation, Comparison between integration and differentiation , Definite Integral - limit of the sum, Different methods of integration, Properties of definite Integral, and some other important question. Students focusing on these topics will surely achieve good marks. 

4. How is Vedantu beneficial for Class 12 students?

Vedantu is a website that focuses on the development of students. This site helps students to avail important study material like notes, sample papers, Exemplars, additional questions, and others. Students can get free study by just registering themselves on Vedantu’s official website i.e. Vedantu.com. Vedantu also offers live Classes for the students to re-understand the concepts taught in Class. Students can clear their doubts through live sessions.