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Class 12 NCERT Mathematics Solutions- Chapter 7– Integrals-Exercise 7.6

Last Updated : 23 Jul, 2025

Chapter 7 of the Class 12 NCERT Mathematics Part II textbook, titled "Integrals," introduces various methods and applications for solving integrals. Exercise 7.6 provides additional practice problems to enhance students' understanding and application of integration techniques.

This section offers detailed solutions for Exercise 7.6 from Chapter 7 of the Class 12 NCERT Mathematics Part II textbook. The exercise includes a range of problems that require applying integration techniques to find integrals of various functions. Solutions are presented step-by-step to assist students in mastering the concepts and methods of integration.

Class 12 NCERT Mathematics Solutions- Exercise 7.6

Question 1: x sinx.

Solution:

Let f(X) = ∫ x sinx dx

Taking x as first function and sin x as second function and integrating by parts, we obtain

f(x) = x ∫sin x dx - {(d(x)/dx) ∫sin x dx} dx

β‡’ f(x) = x(-cosx)-∫1. (-cosx)dx

β‡’ f(x) = x cosx + sinx + C

Question 2: x sin3x

Solution:

Let f(x) = ∫x sin3x dx

Taking x as the first function and sin 3x as the second function and integrating by parts, we obtain

f(x) = x ∫sin3x dx - {(d(x)/dx) ∫sin3x dx} dx

β‡’ f(x) = x (-cos3x βˆ• 3) - ∫1. (cos3x βˆ• 3) dx

β‡’ f(x) = -x cos3xβˆ• 3 + 1βˆ•3 ∫cos3x dx

β‡’ f(x) = -xcos3xβˆ•3 + 1βˆ•9 sin3x + C

Question 3: x2 ex

Solution:

Let f(x) = ∫x2 ex dx

Taking x2 as first function and ex as second function and integrating by parts, we obtain

f(x) = x2 ∫ex dx - {(d(x2)/dx) ∫ex dx} dx

β‡’ f(x) = x2 ex- ∫2x.ex dx

β‡’ f(x) = x2ex - 2 ∫x.ex dx

Again, integrating by parts, we obtain

f(x) = x2ex -2[x.∫exdx-∫{(d(x)/dx). ∫exdx}dx]

β‡’ f(x) = x2ex -2[xex - ∫exdx]

β‡’ f(x) = x2ex-2[xex - ex]

β‡’ f(x) = x2ex -2xex + 2ex + C

β‡’ f(x) = ex (x2-2x+2) + C

Question 4: x logx

Solution:

Let f(x) = ∫x log x dx

Taking log x as first function and x as second function and integrating by parts, we obtain

f(x) = log x ∫x dx - {(d(log x)/dx) ∫x dx} dx

β‡’ f(x) = log x. (x2 βˆ• 2) - ∫1βˆ•x. (x2 βˆ•2) dx

β‡’ f(x) = x2logxβˆ•2- ∫xβˆ•2 dx

β‡’ f(x) = x2logxβˆ•2 - x2βˆ•4 + C

Question 5: x log2x

Solution:

Let f(x) = ∫x log 2x dx

Taking log 2x as first function and x as second function and integrating by parts, we obtain

f(x) = log 2x ∫x dx - {(d (log 2x)βˆ•dx) ∫x dx} dx

β‡’ f(x) = log 2x. (x2 βˆ• 2) - ∫2βˆ•2x. (x2 βˆ•2) dx

β‡’ f(x) = x2log2xβˆ•2 - ∫xβˆ•2 dx

β‡’ f(x) = x2log2xβˆ•2 - x2βˆ•4 + C

Question 6: x2 logx

Solution:

Let f(x) = ∫x2 log x dx

Taking log x as first function and x2 as second function and integrating by parts, we obtain

f(x) = log x ∫x2 dx - {(d (log x) βˆ•dx) ∫x2 dx} dx

β‡’ f(x) = log x. (x3 βˆ• 3) - ∫1βˆ•x. (x3 βˆ•3) dx

β‡’ f(x) = x3logxβˆ•3 - ∫x3βˆ• 3 dx

β‡’ f(x) = x3logxβˆ•3 - x3 βˆ• 9 + C

Question 7: x sin-1x.

Solution:

Let f(x) = ∫x sin-1x dx

Taking sin-1x as first function and x as second function and integrating by parts, we obtain

f(x) = sin-1x ∫ x dx - ∫ {(d(sin-1xβˆ•dx) ∫x dx} dx

β‡’ f(x) = sin-1x (x2/2) - ∫ 1βˆ• √(1-x2 ).x2βˆ•2 dx

β‡’ f(x) = x2sin-1x βˆ•2 + 1βˆ•2 ∫-x βˆ• √(1-x2 ) dx

β‡’ f(x) = x2 sin-1xβˆ•2 + 1βˆ•2 ∫ {1-x2 βˆ• √(1-x2 )- 1βˆ•βˆš(1-x2)} dx

β‡’ f(x) = x2 sin-1xβˆ•2 + 1βˆ•2 ∫ {√1-x2 - 1βˆ•βˆš(1-x2)} dx

β‡’ f(x) = x2 sin-1xβˆ•2 + 1βˆ•2 {∫ √1-x2 dx - 1 βˆ•βˆš(1-x2) dx}

β‡’ f(x) = x2sin-1xβˆ•2 + 1βˆ•2{xβˆ•2. √1-x2 + 1βˆ•2sin-1x - sin-1x} +C

β‡’ f(x) = x2sin-1xβˆ•2 + xβˆ•4. √1-x2 + 1βˆ•4sin-1x - 1βˆ•2sin-1x +C

β‡’ f(x) = 1βˆ•4(2x2-1) sin-1x + xβˆ•4. √1-x2 + C

Question 8: x tan-1x

Solution:

Let f(x) = ∫ x tan-1x dx

Taking tan-1x as first function and x as second function and integrating by parts, we obtain

f(x) = tan-1x ∫ x dx - ∫ {(d(tan-1xβˆ•dx) ∫x dx} dx

β‡’ f(x) = tan-1x (x2βˆ•2) - ∫ 1βˆ• (1+x). x2βˆ•2 dx

β‡’ f(x) = x2 tan-1x βˆ• 2 - 1βˆ•2∫ 1βˆ• (1+x2 ) dx

β‡’ f(x) = x2 tan-1x βˆ• 2 - 1βˆ•2∫{(x2+1) βˆ• (1+x2) - 1βˆ• (1+x2)} dx

β‡’ f(x) = x2 tan-1x βˆ• 2 - 1βˆ•2∫ (1- 1βˆ• (1+x2) dx

β‡’ f(x) = x2 tan-1x βˆ• 2 - 1βˆ•2 (x-tan-1 x) + C

β‡’ f(x) = x2 tan-1x βˆ• 2 - xβˆ•2 + 1βˆ•2 tan-1x + C

Question 9: x cos-1x

Solution:

Let f(x) = ∫x cos-1x dx

Taking cos-1x as first function and x as second function and integrating by parts, we obtain

f(x) = cos-1x ∫ x dx - ∫ {(d(cos-1x)βˆ•dx.∫x dx} dx

β‡’ f(x) = cos-1x (x2/2) - ∫ -1βˆ• √1-x2 .x2βˆ•2 dx

β‡’ f(x) = x2 cos-1x βˆ•2 - 1βˆ•2 ∫1-x2-1 βˆ• √(1-x2 ) dx

β‡’ f(x) = x2 cos-1xβˆ•2 - 1βˆ•2 ∫ {√(1-x2 ) +( -1βˆ•βˆš1-x2)} dx

β‡’ f(x) = x2 cos-1xβˆ•2 - 1βˆ•2 ∫ √1-x2 dx - 1βˆ•2 ∫(-1βˆ•βˆš1-x2) dx

β‡’ f(x) = x2 cos-1xβˆ•2 - 1βˆ•2 I1 - 1βˆ•2 cos-1x -------------(1)

Where I1 = ∫√1-x2 dx

I1 = x√1-x2 - ∫d(√1-x2)βˆ•dx ∫x dx

β‡’ I1 = x√1-x2 - ∫d(-2xβˆ•2√1-x2 .x dx

β‡’ I1 = x√1-x2 - ∫-x2βˆ•βˆš1-x2 dx

β‡’ I1 = x√1-x2 - ∫1-x2-1 βˆ• √(1-x2) dx

β‡’ I1 = x√1-x2 - { ∫ √1-x2 dx + ∫(-dxβˆ•βˆš1-x2}

β‡’ I1 = x√1-x2- {I1 + cos-1x}

β‡’ 2I1 = x√1-x2 - cos-1x

β‡’ I1 = xβˆ•2 .√1-x2 -1βˆ•2 cos-1x

Substituting in (1), we obtain

f(x) = x2cos-1xβˆ•2 - 1βˆ•2 (xβˆ•2 .√1-x2 -1βˆ•2 cos-1x) - 1βˆ•2 cos-1x

β‡’ f(x) = (2x-1)βˆ•4 cos-1x - xβˆ•4 √1-x2 + C

Question 10: (sin-1x)2

Solution:

Let f(x) = ∫(sin-1x)2 dx

Taking (sin-1x)2as first function and 1 as second function and integrating by parts, we obtain

f(x) = (sin-1x)2∫ 1dx - ∫ {(d(sin-1x)2βˆ•dx. ∫1 dx} dx

β‡’ f(x) = (sin-1x). x - ∫ 2.sin-1xβˆ•βˆš1-x2 . x dx

β‡’ f(x) = x(sin-1x) 2 + ∫sin-1x.(-2xβˆ• √1-x2)dx

β‡’ f(x) = x(sin-1x) 2+[sin-1x∫-2xβˆ•βˆš1-x2 dx - ∫ {❴d(sin-1x)βˆ•dx❡∫-2x βˆ•βˆš1-x2 dx}dx]

β‡’ f(x) =x(sin-1x) 2+[ sin-1x .2√1-x2 - ∫1βˆ•βˆš1-x2 .2√1-x2 dx ]

β‡’ f(x) =x(sin-1x)2+ 2√1-x2 sin-1x - ∫2dx

β‡’ f(x) =x(sin-1x)2+ 2√1-x2 .sin-1x -2x + C

Question 11: (x cos-1x) / √1-x2

Solution:

Let f(x) = ∫(x cos-1x) / √1-x2 dx

we are multiplying -1/2 in numerator and dementor then.

f(x) = -1βˆ•2 ∫(-2x cos-1x) βˆ•βˆš1-x2 dx

Taking (cos-1x) as first function and {-2xβˆ•βˆš1-x2 ❡ as second function and integrating by parts, we obtain

f(x) = -1βˆ•2 [ cos-1x∫-2xβˆ•βˆš1-x dx - ∫ {❲d(cos-1x)βˆ•dx❳∫-2x√1-x2 dx}dx]

β‡’ f(x) = -1βˆ•2 [cos-1x.2√1-x2 - ∫-1βˆ•βˆš1-x2 2√1-x2 dx]

β‡’ f(x) = -1βˆ•2 [2√1-x2 cos-1x+ ∫2 dx]

β‡’ f(x) = -1βˆ•2 [2√1-x2 cos-1x+ 2x] + C

β‡’ f(x) = - [√1-x2 cos-1x+ x] + C

Question 12: x sec2x

Solution:

Let f(x) = ∫x sec2x dx

Taking x as first function and sec2x as second function and integrating by parts, we obtain

f(x) = x ∫sec2x dx - {(d (x) βˆ•dx) ∫sec2x dx} dx

β‡’ f(x) = xtanx - ∫1.tanx dx

β‡’ f(x) = x tanx + log|cosx| + C

Question 13: tan-1x

Solution:

Let f(x) = ∫tan-1x dx

Taking tan-1x as first function and 1 as second function and integrating by parts, we obtain

f(x) = tan-1x∫1dx - ∫ {[d❲tan-1xβ³βˆ•dx] ∫1. dx}dx

β‡’ f(x) = tan-1x .x - ∫1βˆ•1+x2 .x dx

β‡’ f(x) = x tan-1x - 1βˆ•2 ∫2xβˆ•1+x2 dx

β‡’ f(x) = x tan-1x - 1βˆ•2 log|1+x2| +C

β‡’ f(x) = x tan-1x -1βˆ•2 log(1+x2) + C

Question 14: x (logx)2

Solution:

Let f(x) = ∫ x (logx)2 dx

Taking (logx)2 as first function and x as second function and integrating by parts, we obtain

f(x) = (logx)2 ∫x dx - ∫[{{d(logx)βˆ•dx}2}∫xdx]dx

β‡’ f(x) = xβˆ•2 (logx)2 - [∫2logx 1βˆ•x . x2βˆ•2 dx]

β‡’ f(x) = x2βˆ•2 (logx)2 - ∫x logx dx

Again, integration by parts, we obtain.

f(x) = x2βˆ•2 ❲logx❳2 - [logx ∫x dx - ∫{❴d(logx)βˆ•dx}∫xdx❡dx]

β‡’ f(x) = x2βˆ•2 ❲logx❳2 - [x2βˆ•2 - logx - ∫1βˆ•x .x2βˆ•2 dx]

β‡’ f(x) = x2βˆ•2 ❲logx❳2 - x2βˆ•2 .logx + 1βˆ•2∫xdx

β‡’ f(x) = x2βˆ•2 ❲logx❳2 - x2βˆ•2 .logx + x2βˆ•4 + C

Question 15: (x2+1) logx

Solution:

Let f(x) = ∫ (x2+1) logx dx

β‡’ f(x) = ∫ x2 logx dx + ∫ logx dx

Let f(x) = I1 + I2 ............................. (1)

where I1 = ∫ x2 logx dx and I2 = ∫ logx dx

I1 = ∫ x2 logx dx

Taking (logx) as first function and x2 as second function and integrating by parts, we obtain

β‡’ I1 = logx - ∫ x2dx - ∫{❴d(logx)βˆ•dx❡∫x2dx} dx

β‡’ I1 = logx .x3βˆ•3 - ∫1βˆ•x . x3βˆ•3 dx

β‡’ I1 = x3βˆ•3 logx - 1βˆ•3(∫x dx)

β‡’ I1 = x3βˆ•3 logx - x3βˆ•9 + C1 ........................ (2)

I2 = ∫ logx dx

Taking log x as first function and 1 as second function and integrating by parts, we obtain

f(x) = log x ∫1 dx - {(d(log x)/dx) ∫1 dx} dx

β‡’ f(x)= log x.x - ∫1βˆ•x. x dx

β‡’ f(x) = x. logxβˆ•2 - ∫1 dx

β‡’ f(x) = x. logxβˆ•2 - x + C2 ......................(3)

Using equation (2) and (3) in (1), we obtain

f(x) = x3βˆ•3 logx - x3βˆ•9 + C1 + x. logxβˆ•2 - x + C2

β‡’ f(x)= x3βˆ•3 logx - x3βˆ•9 + x. logxβˆ•2 - x + C 1+ C2

β‡’ f(x)= (x3βˆ•3 + x) logx - x3βˆ•9 - x + C

Question 16: ex(sinx + cosx)

Solution:

Let f(x) = ∫ e(sinx+cosx) dx

Let g(x) = sinx

g'(x) = cosx

f(x) = ∫ ex{g(x) + g'(x) } dx

It is known that, ∫ ex{ g(x) + g'(x) } dx = ex g(x) + C

So, f(x) = ex sinx + C

Question 17: x ex βˆ•(1+x)2

Solution:

Let f(x) = ∫ x ex βˆ•(1+x)2 dx

β‡’ f(x)= ∫ ex{x βˆ•(1+x)2} dx

β‡’ f(x)= ∫ ex{(1+x-1) βˆ• (1+x)2} dx

β‡’ f(x)= ∫ e {1βˆ• (1+x) - 1βˆ• (1+x)2} dx

Let f(x) = 1βˆ• (1+x)

f'(x) = -1βˆ• (1+x)2

β‡’ f(x) = ∫ x ex βˆ•(1+x)2 dx = ∫ ex{f(x) + f'(x) }dx

It is known that ∫ ex{f(x) + f'(x) }dx = ex f(x) + C

So, ∫ x ex βˆ•(1+x)2 dx = ex βˆ• (1+x) C

Question 18: ex{(1+sinx) βˆ• (1+cosx)}

Solution:

Let I = ex{(1+sinx)βˆ•(1+cosx)}

β‡’ I =ex{(sin2xβˆ•2+cos2xβˆ•2+2 sinxβˆ•2 cosxβˆ•2) βˆ• (2cos2xβˆ•2)}

β‡’ I = ex{(sinxβˆ•2+cosxβˆ•2)2 βˆ• (2cos2xβˆ•2)}

β‡’ I = 1βˆ•2 ex{(sinxβˆ•2+cosxβˆ•2) βˆ• (cosxβˆ•2)}2

β‡’ I = 1βˆ•2 ex {tanxβˆ•2 + 1}2

β‡’ I = 1βˆ•2 ex{1 + tan2xβˆ•2 + 2tanxβˆ•2}

β‡’ I = 1βˆ•2 ex{secxβˆ•2 + 2tanxβˆ•2}

β‡’ I = ex(1+sinx)dx βˆ• (1+cosx) = ex{1βˆ•2 sec2xβˆ•2 + tanxβˆ•2} ----------------- (1)

Let tanxβˆ•2 = f(x) or f'(x) = 1βˆ•2 sec2xβˆ•2

It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C

from equation (1), we obtain,

∫ex{(1+sinx)βˆ•(1+cosx)}dx = ex tanxβˆ•2 + C

Question 19: ex{1βˆ•x - 1βˆ•x2}

Solution:

Let f(x) = ∫ex{1βˆ•x - 1βˆ•x2} dx

Also Let 1βˆ•x = f(x) or f'(x) = - 1βˆ•x2

It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C

So, f(x) = ex βˆ• x + C

Question 20: (x-3) ex βˆ• (x-1)3

Solution:

Let f(x) = ∫ ex{ (x-3) βˆ• (x-1)3}dx

β‡’ f(x) = ∫ ex{ (x-1-2) βˆ• (x-1)3}dx

β‡’ f(x) = ∫ ex{ 1βˆ• (x-1)2 - 2 βˆ• (x-1)3}dx

β‡’ f(x) = 1 βˆ• (x-1)2 or f'(x) = -2 βˆ• (x-1)3

It is known that ∫ ex{f(x) + f'(x) } dx = ex f(x) + C

So, ∫ ex{ (x-3) βˆ• (x-1)3}dx = ex βˆ• {x-1}2 + C

Question 21: e2x sinx

Solution:

Let f(x) = ∫ e2x sinx dx -------------- (1)

Integrating by parts, we obtain

f(x) = sinx ∫ e2x dx - ∫ { ❴ d(sinx)βˆ•dx❡ ∫ e2x dx} dx

β‡’ f(x) = sinx . e2xβˆ•2 - ∫ cosx e2xβˆ•2 dx

β‡’ f(x) = 1βˆ•2 e2x sinx - 1βˆ•2 ∫ e2x cosx dx

Again, Integrating by parts, we obtain

f(x) = 1βˆ•2 e2x sinx - 1βˆ•2 [cosx ∫ e2x dx - ∫ {❴dβˆ•dx cosx❡∫e2x dx} dx]

β‡’ f(x) = 1βˆ•2 e2x sinx - 1βˆ•2 [cosx e2x βˆ• 2 - ∫ ❴-sinx❡ e2x βˆ•2 dx

β‡’ f(x) = 1βˆ•2 e2x sinx - 1βˆ•2 [(cosx e2x ) βˆ• 2 + ∫ sinx e2x βˆ•2 dx

β‡’ f(x) = 1βˆ•2 e2x sinx - (e2x cosx) βˆ• 4 - 1βˆ•4f(x) ----------from (1)

β‡’ f(x) + 1βˆ•4f(x) = 1βˆ•2 e2x sinx - (e2x cosx) βˆ• 4

β‡’ 5/4 f(x) = (e2x sinx)1βˆ•2 - (e2x cosx) βˆ• 4

β‡’ f(x) = e2x βˆ•5 [ 2 sinx - cosx] + C

Question 22: sin-1(2xβˆ•(1+x2)

Solution:

sin-1(2xβˆ•(1+x2)

Let x = tanΞΈ or dx = sec2ΞΈ dΞΈ

So, sin-1(2xβˆ•(1+x2) = sin-1(2 tanΞΈβˆ•(1+tan2ΞΈ) = sin-1(sin2ΞΈ) = 2ΞΈ

Integrating by parts, we obtain

2[ΞΈ.∫sec2ΞΈ dΞΈ - ∫{❴dΞΈβˆ• dθ❡sec2ΞΈ dΞΈ} dΞΈ]

= 2[θ.tanθ - ∫tanθ dθ]

= 2[ΞΈ.tanΞΈ - log|cosΞΈ|] + C

= 2[x.tan-1x - log|1√(1+x2|] + C

= 2xtan-1x + 2 log(1+x2)1βˆ•2 + C

= 2x tan-1x + 2 [-1βˆ•2 log(1+x2) ] + C

= 2x tan-1x - log(1+x2) + C

Choose the correct answer in Exercises 23 and 24.

Question 23: ∫ x2ex^3 dx equals

(A) 1βˆ•3 .eX^3 + C

(B) 1βˆ•3 . eX^3 + C

(C) 1βˆ•2 .eX^3 + C

(D) 1βˆ•3 . eX^3 + C

Correct answer is A.

Question 24 :∫ ex secx(1+tanx) dx equals.

(A) ex cosx + C

(B) ex secx + C

(C) ex sinx + C

(D) ex tanx + C

Correct answer is (B) ex secx + C.

Also Read:

Conclusion

Chapter 7 of the Class 12 NCERT Mathematics Part II textbook, "Integrals," covers techniques for solving integrals, including substitution, integration by parts, and partial fractions. Exercise 7.6 provides practice problems to apply these methods. Solutions are detailed to help students effectively understand and solve various integral problems.

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