🔥 Jadwal & Hasil: maths/class ncert solutions mathematics part ii chapter integrals exercise

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
⇒ f(x) = x(-cosx)-∫1. (-cosx)dx
⇒ 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
⇒ f(x) = x (-cos3x ∕ 3) - ∫1. (cos3x ∕ 3) dx
⇒ f(x) = -x cos3x∕ 3 + 1∕3 ∫cos3x dx
⇒ f(x) = -xcos3x∕3 + 1∕9 sin3x + C

Question 3: x²e^x

Solution:

Let f(x) = ∫x² e^x dx
Taking x² as first function and e^x as second function and integrating by parts, we obtain
f(x) = x² ∫e^x dx - {(d(x²)/dx) ∫e^x dx} dx
⇒ f(x) = x²e^x- ∫2x.e^x dx
⇒ f(x) = x²e^x - 2 ∫x.e^xdx
Again, integrating by parts, we obtain
f(x) = x²e^x-2[x.∫e^xdx-∫{(d(x)/dx). ∫e^xdx}dx]
⇒ f(x) = x²ex -2[xe^x - ∫e^xdx]
⇒ f(x) = x²e^x-2[xe^x- e^x]
⇒ f(x) = x²e^x-2xe^x + 2e^x+ C
⇒ f(x) = e^x(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
⇒ f(x) = log x. (x² ∕ 2) - ∫1∕x. (x² ∕2) dx
⇒ f(x) = x²logx∕2- ∫x∕2 dx
⇒ f(x) = x²logx∕2 - x²∕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)∕dx) ∫x dx} dx
⇒ f(x) = log 2x. (x² ∕ 2) - ∫2∕2x. (x² ∕2) dx
⇒ f(x) = x²log2x∕2 - ∫x∕2 dx
⇒ f(x) = x²log2x∕2 - x2∕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) ∕dx) ∫x² dx} dx
⇒ f(x) = log x. (x³ ∕ 3) - ∫1∕x. (x³ ∕3) dx
⇒ f(x) = x³logx∕3 - ∫x³∕ 3 dx
⇒ f(x) = x³logx∕3 - x³ ∕ 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^-1x∕dx) ∫x dx} dx
⇒ f(x) = sin^-1x (x²/2) - ∫ 1∕ √(1-x² ).x²∕2 dx
⇒ f(x) = x²sin^-1x ∕2 + 1∕2 ∫-x ∕ √(1-x² ) dx
⇒ f(x) = x²sin^-1x∕2 + 1∕2 ∫ {1-x²∕ √(1-x² )- 1∕√(1-x²)} dx
⇒ f(x) = x² sin^-1x∕2 + 1∕2 ∫ {√1-x² - 1∕√(1-x²)} dx
⇒ f(x) = x² sin-1x∕2 + 1∕2 {∫ √1-x²dx - 1 ∕√(1-x²) dx}
⇒ f(x) = x²sin^-1x∕2 + 1∕2{x∕2. √1-x² + 1∕2sin^-1x - sin^-1x} +C
⇒ f(x) = x²sin^-1x∕2 + x∕4. √1-x² + 1∕4sin^-1x - 1∕2sin^-1x +C
⇒ f(x) = 1∕4(2x²-1) sin^-1x + x∕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^-1x∕dx) ∫x dx} dx
⇒ f(x) = tan^-1x (x²∕2) - ∫ 1∕ (1+x). x²∕2 dx
⇒ f(x) = x² tan^-1x ∕ 2 - 1∕2∫ 1∕ (1+x² ) dx
⇒ f(x) = x²tan^-1x ∕ 2 - 1∕2∫{(x²+1) ∕ (1+x²) - 1∕ (1+x²)} dx
⇒ f(x) = x² tan^-1x ∕ 2 - 1∕2∫ (1- 1∕ (1+x²) dx
⇒ f(x) = x²tan^-1x ∕ 2 - 1∕2 (x-tan^-1x) + C
⇒ f(x) = x² tan^-1x ∕ 2 - x∕2 + 1∕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)∕dx.∫x dx} dx
⇒ f(x) = cos^-1x (x²/2) - ∫ -1∕ √1-x² .x²∕2 dx
⇒ f(x) = x²cos^-1x ∕2 - 1∕2 ∫1-x²-1 ∕ √(1-x² ) dx
⇒ f(x) = x² cos^-1x∕2 - 1∕2 ∫ {√(1-x² ) +( -1∕√1-x²)} dx
⇒ f(x) = x² cos^-1x∕2 - 1∕2 ∫ √1-x² dx - 1∕2 ∫(-1∕√1-x²) dx
⇒ f(x) = x² cos^-1x∕2 - 1∕2 I₁ - 1∕2 cos^-1x -------------(1)
Where I₁= ∫√1-x² dx
I₁ = x√1-x² - ∫d(√1-x²)∕dx ∫x dx
⇒ I₁ = x√1-x² - ∫d(-2x∕2√1-x² .x dx
⇒ I₁ = x√1-x² - ∫-x²∕√1-x² dx
⇒ I₁ = x√1-x² - ∫1-x²-1 ∕ √(1-x²) dx
⇒ I₁ = x√1-x² - { ∫ √1-x² dx + ∫(-dx∕√1-x²}
⇒ I₁ = x√1-x²- {I₁ + cos^-1x}
⇒ 2I₁ = x√1-x² - cos^-1x
⇒ I₁ = x∕2 .√1-x² -1∕2 cos^-1x
Substituting in (1), we obtain
f(x) = x²cos^-1x∕2 - 1∕2 (x∕2 .√1-x² -1∕2 cos^-1x) - 1∕2 cos^-1x
⇒ f(x) = (2x-1)∕4 cos^-1x - x∕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)²∕dx. ∫1 dx} dx
⇒ f(x) = (sin^-1x). x - ∫ 2.sin^-1x∕√1-x² . x dx
⇒ f(x) = x(sin^-1x) ² + ∫sin^-1x.(-2x∕ √1-x²⁾dx
⇒ f(x) = x(sin^-1x) ²+[sin^-1x∫-2x∕√1-x² dx - ∫ {❴d(sin^-1x)∕dx❵∫-2x ∕√1-x² dx}dx]
⇒ f(x) =x(sin^-1x) ²+[ sin^-1x .2√1-x² - ∫1∕√1-x² .2√1-x² dx ]
⇒ f(x) =x(sin^-1x)²+ 2√1-x² sin^-1x - ∫2dx
⇒ 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-x²dx
we are multiplying -1/2 in numerator and dementor then.
f(x) = -1∕2 ∫(-2x cos^-1x) ∕√1-x² dx
Taking (cos^-1x) as first function and {-2x∕√1-x² ❵ as second function and integrating by parts, we obtain
f(x) = -1∕2 [ cos^-1x∫-2x∕√1-x dx - ∫ {❲d(cos^-1x)∕dx❳∫-2x√1-x² dx}dx]
⇒ f(x) = -1∕2 [cos^-1x.2√1-x² - ∫-1∕√1-x² 2√1-x² dx]
⇒ f(x) = -1∕2 [2√1-x² cos^-1x+ ∫2 dx]
⇒ f(x) = -1∕2 [2√1-x² cos^-1x+ 2x] + C
⇒ 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) ∕dx) ∫sec²x dx} dx
⇒ f(x) = xtanx - ∫1.tanx dx
⇒ 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^-1x❳∕dx] ∫1. dx}dx
⇒ f(x) = tan^-1x .x - ∫1∕1+x² .x dx
⇒ f(x) = x tan^-1x - 1∕2 ∫2x∕1+x² dx
⇒ f(x) = x tan^-1x - 1∕2 log|1+x²| +C
⇒ f(x) = x tan^-1x -1∕2 log(1+x²) + C

Question 14: x (logx)²

Solution:

Let f(x) = ∫ x (logx)²dx
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)∕dx}²}∫xdx]dx
⇒ f(x) = x∕2 (logx)² - [∫2logx 1∕x . x²∕2 dx]
⇒ f(x) = x²∕2 (logx)² - ∫x logx dx
Again, integration by parts, we obtain.
f(x) = x²∕2 ❲logx❳² - [logx ∫x dx - ∫{❴d(logx)∕dx}∫xdx❵dx]
⇒ f(x) = x²∕2 ❲logx❳² - [x²∕2 - logx - ∫1∕x .x²∕2 dx]
⇒ f(x) = x²∕2 ❲logx❳² - x²∕2 .logx + 1∕2∫xdx
⇒ f(x) = x²∕2 ❲logx❳² - x²∕2 .logx + x²∕4 + C

Question 15: (x²+1) logx

Solution:

Let f(x) = ∫ (x²+1) logx dx
⇒ 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
⇒ I₁ = logx - ∫ x²dx - ∫{❴d(logx)∕dx❵∫x²dx} dx
⇒ I₁ = logx .x³∕3 - ∫1∕x . x³∕3 dx
⇒ I₁ = x³∕3 logx - 1∕3(∫x dx)
⇒ I₁ = x³∕3 logx - x³∕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
⇒ f(x)= log x.x - ∫1∕x. x dx
⇒ f(x) = x. logx∕2 - ∫1 dx
⇒ f(x) = x. logx∕2 - x + C₂ ......................(3)
Using equation (2) and (3) in (1), we obtain
f(x) = x³∕3 logx - x³∕9 + C1 + x. logx∕2 - x + C2
⇒ f(x)= x³∕3 logx - x³∕9 + x. logx∕2 - x + C ₁+ C2
⇒ f(x)= (x³∕3 + x) logx - x³∕9 - x + C

Question 16: e^x(sinx + cosx)

Solution:

Let f(x) = ∫ e(sinx+cosx) dx
Let g(x) = sinx
g^'(x) = cosx
f(x) = ∫ e^x{g(x) + g^'(x) } dx
It is known that, ∫ e^x{ g(x) + g'(x) } dx = e^xg(x) + C
So, f(x) = e^x sinx + C

Question 17: x e^x ∕(1+x)²

Solution:

Let f(x) = ∫ x e^x ∕(1+x)² dx
⇒ f(x)= ∫ e^x{x ∕(1+x)²} dx
⇒ f(x)= ∫ e^x{(1+x-1) ∕ (1+x)²} dx
⇒ f(x)= ∫ e {1∕ (1+x) - 1∕ (1+x)²} dx
Let f(x) = 1∕ (1+x)
f^'(x) = -1∕ (1+x)²
⇒ f(x) = ∫ x e^x ∕(1+x)² dx = ∫ e^x{f(x) + f'(x) }dx
It is known that ∫ e^x{f(x) + f'(x) }dx = e^x f(x) + C
So, ∫ x e^x ∕(1+x)² dx = e^x ∕ (1+x) C

Question 18: e^x{(1+sinx) ∕ (1+cosx)}

Solution:

Let I = e^x{(1+sinx)∕(1+cosx)}
⇒ I =e^x{(sin²x∕2+cos²x∕2+2 sinx∕2 cosx∕2) ∕ (2cos²x∕2)}
⇒ I = e^x{(sinx∕2+cosx∕2)² ∕ (2cos²x∕2)}
⇒ I = 1∕2 e^x{(sinx∕2+cosx∕2) ∕ (cosx∕2)}²
⇒ I = 1∕2 e^x {tanx∕2 + 1}²
⇒ I = 1∕2 e^x{1 + tan²x∕2 + 2tanx∕2}
⇒ I = 1∕2 e^x{secx∕2 + 2tanx∕2}
⇒ I = e^x(1+sinx)dx ∕ (1+cosx) = e^x{1∕2 sec²x∕2 + tanx∕2} ----------------- (1)
Let tanx∕2 = f(x) or f'(x) = 1∕2 sec²x∕2
It is known that ∫ e^x{f(x) + f'(x) } dx = e^xf(x) + C
from equation (1), we obtain,
∫e^x{(1+sinx)∕(1+cosx)}dx = e^x tanx∕2 + C

Question 19: e^x{1∕x - 1∕x²}

Solution:

Let f(x) = ∫e^x{1∕x - 1∕x²} dx
Also Let 1∕x = f(x) or f'(x) = - 1∕x²
It is known that ∫ e^x{f(x) + f'(x) } dx = e^xf(x) + C
So, f(x) = e^x ∕ x + C

Question 20: (x-3) e^x ∕ (x-1)³

Solution:

Let f(x) = ∫ e^x{ (x-3) ∕ (x-1)³}dx
⇒ f(x) = ∫ e^x{ (x-1-2) ∕ (x-1)³}dx
⇒ f(x) = ∫ e^x{ 1∕ (x-1)² - 2 ∕ (x-1)³}dx
⇒ f(x) = 1 ∕ (x-1)² or f'(x) = -2 ∕ (x-1)³
It is known that ∫ e^x{f(x) + f'(x) } dx = e^x f(x) + C
So, ∫ e^x{ (x-3) ∕ (x-1)³}dx = e^x ∕ {x-1}² + C

Question 21: e^2x sinx

Solution:

Let f(x) = ∫ e^2x sinx dx -------------- (1)
Integrating by parts, we obtain
f(x) = sinx ∫ e^2x dx - ∫ { ❴ d(sinx)∕dx❵ ∫ e^2xdx} dx
⇒ f(x) = sinx . e^2x∕2 - ∫ cosx e^2x∕2 dx
⇒ f(x) = 1∕2 e^2x sinx - 1∕2 ∫ e^2x cosx dx
Again, Integrating by parts, we obtain
f(x) = 1∕2 e^2xsinx - 1∕2 [cosx ∫ e^2x dx - ∫ {❴d∕dx cosx❵∫e^2x dx} dx]
⇒ f(x) = 1∕2 e^2x sinx - 1∕2 [cosx e^2x ∕ 2 - ∫ ❴-sinx❵ e^2x ∕2 dx
⇒ f(x) = 1∕2 e^2x sinx - 1∕2 [(cosx e^2x ) ∕ 2 + ∫ sinx e^2x ∕2 dx
⇒ f(x) = 1∕2 e^2x sinx - (e^2x cosx) ∕ 4 - 1∕4f(x) ----------from (1)
⇒ f(x) + 1∕4f(x) = 1∕2 e^2x sinx - (e^2x cosx) ∕ 4
⇒ 5/4 f(x) = (e^2x sinx)1∕2 - (e2x cosx) ∕ 4
⇒ f(x) = e^2x ∕5 [ 2 sinx - cosx] + C

Question 22: sin^-1(2x∕(1+x²)

Solution:

sin^-1(2x∕(1+x²)
Let x = tanθ or dx = sec²θ dθ
So, sin^-1(2x∕(1+x²) = sin^-1(2 tanθ∕(1+tan²θ) = sin^-1(sin2θ) = 2θ
Integrating by parts, we obtain
2[θ.∫sec²θ dθ - ∫{❴dθ∕ dθ❵sec²θ dθ} dθ]
= 2[θ.tanθ - ∫tanθ dθ]
= 2[θ.tanθ - log|cosθ|] + C
= 2[x.tan^-1x - log|1√(1+x²|] + C
= 2xtan^-1x + 2 log(1+x²)^1∕2 + C
= 2x tan^-1x + 2 [-1∕2 log(1+x²) ] + C
= 2x tan^-1x - log(1+x²) + C

Choose the correct answer in Exercises 23 and 24.

Question 23: ∫ x²e^{x^3} dx equals

(A) 1∕3 .e^{X^3} + C

(B) 1∕3 . e^{X^3} + C

(C) 1∕2 .e^{X^3} + C

(D) 1∕3 . e^{X^3} + C

Correct answer is A.

Question 24 :∫ e^x secx(1+tanx) dx equals.

(A) e^x cosx + C

(B) e^x secx + C

(D) e^x tanx + C

Correct answer is (B) e^x secx + C.

Also Read:

Integrals
Properties of Definite Integrals
Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.5
Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.7

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.

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

Class 12 NCERT Mathematics Solutions- Exercise 7.6

Question 1: x sinx.

Question 2: x sin3x

Question 3: x²e^x

Question 4: x logx

Question 5: x log2x

Question 6: x² logx

Question 7: x sin^-1x.

Question 8: x tan^-1x

Question 9: x cos^-1x

Question 10: (sin^-1x)²

Question 11: (x cos^-1x) / √1-x²

Question 12: x sec²x

Question 13: tan^-1x

Question 14: x (logx)²

Question 15: (x²+1) logx

Question 16: e^x(sinx + cosx)

Question 17: x e^x ∕(1+x)²

Question 18: e^x{(1+sinx) ∕ (1+cosx)}

Question 19: e^x{1∕x - 1∕x²}

Question 20: (x-3) e^x ∕ (x-1)³

Question 21: e^2x sinx

Question 22: sin^-1(2x∕(1+x²)

Choose the correct answer in Exercises 23 and 24.

Question 23: ∫ x²e^{x^3} dx equals

Question 24 :∫ e^x secx(1+tanx) dx equals.

Conclusion

Explore

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

Class 12 NCERT Mathematics Solutions- Exercise 7.6

Question 1: x sinx.

Question 2: x sin3x

Question 3: x2 ex

Question 4: x logx

Question 5: x log2x

Question 6: x2 logx

Question 7: x sin-1x.

Question 8: x tan-1x

Question 9: x cos-1x

Question 10: (sin-1x)2

Question 11: (x cos-1x) / √1-x2

Question 12: x sec2x

Question 13: tan-1x

Question 14: x (logx)2

Question 15: (x2+1) logx

Question 16: ex(sinx + cosx)

Question 17: x ex ∕(1+x)2

Question 18: ex{(1+sinx) ∕ (1+cosx)}

Question 19: ex{1∕x - 1∕x2}

Question 20: (x-3) ex ∕ (x-1)3

Question 21: e2x sinx

Question 22: sin-1(2x∕(1+x2)

Choose the correct answer in Exercises 23 and 24.

Question 23: ∫ x2ex^3 dx equals

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

Conclusion

Explore

Question 3: x²e^x

Question 6: x² logx

Question 7: x sin^-1x.

Question 8: x tan^-1x

Question 9: x cos^-1x

Question 10: (sin^-1x)²

Question 11: (x cos^-1x) / √1-x²

Question 12: x sec²x

Question 13: tan^-1x

Question 14: x (logx)²

Question 15: (x²+1) logx

Question 16: e^x(sinx + cosx)

Question 17: x e^x ∕(1+x)²

Question 18: e^x{(1+sinx) ∕ (1+cosx)}

Question 19: e^x{1∕x - 1∕x²}

Question 20: (x-3) e^x ∕ (x-1)³

Question 21: e^2x sinx

Question 22: sin^-1(2x∕(1+x²)

Question 23: ∫ x²e^{x^3} dx equals

Question 24 :∫ e^x secx(1+tanx) dx equals.