Students often turn to Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.11 to clarify doubts and improve problem-solving skills.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.11

Very Short answer type questions (1 to 3) :

Evaluate the following (1 to 22) integrals

Question 1.
(i) ∫ x ex dx
(ii) ∫ x sin x dx (NCERT)
Solution:
(i) ∫ x ex dx = x . ex – ∫ 1 . ex dx
[Integrating by parts]
= x ex – ex + C
= (x – 1) ex + C

(ii) ∫ x sin x dx = – x cos x – ∫ 1 . (- cos x) dx
= – x cos x + sin x + C

Question 2.
(i) ∫ x sec2 x dx (NCERT)
(ii) ∫ x sin 3x dx (NCERT)
Solution:
(i) ∫ x sec2 x dx
= x tan x – ∫ 1 . tan x dx
= x tan x – ∫ $$\frac{\sin x}{\cos x}$$ dx
= x log x + log |cos x| + C

(ii) ∫ x sin 3x dx
= x $$\left(-\frac{\cos 3 x}{3}\right)$$ + ∫ 1 . $$\frac{\cos 3 x}{3}$$ dx
= – x $$\frac{\cos 3 x}{3}$$ + $$\frac{\sin 3 x}{9}$$ + C

Question 3.
(i) ∫ x e2x dx
(ii) ∫ x cos 2x dx
Solution:
(i) ∫ x e2x dx = $$\frac{x e^{2 x}}{2}-\int 1 \cdot \frac{e^{2 x}}{2} d x$$
= $$\frac{x e^{2 x}}{2}-\frac{e^{2 x}}{4}$$
= $$\frac{1}{4}$$ (2x – 1) e2x + C

(ii) ∫ x cos 2x dx = $$x \frac{\sin 2 x}{2}-\int 1 \cdot \frac{\sin 2 x}{2} d x$$
= $$\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}$$ + C

Question 4.
(i) ∫ x sec2 x tan x dx
(ii) ∫ (elog x + sin x) cos x dx
Solution:
(i) ∫ x sec2 x tan x dx
= x . $$\frac{\tan ^2 x}{2}$$ – ∫ 1 . $$\frac{\tan ^2 x}{2}$$ dx
[∵ ∫ sec2 x tan x dx = ∫ tan x . sec2 x dx
= $$\frac{\tan ^2 x}{2}$$ ;
Since ∫ [f(x)]n f'(x) dx = $$\frac{[f(x)]^{n+1}}{n+1}$$, n ≠ – 1]
= $$\frac{x \tan ^2 x}{2}$$ – $$\frac{1}{2}$$ ∫ (sec2 x – 1) dx
= $$\frac{x \tan ^2 x}{2}$$ – $$\frac{1}{2}$$ tan x + $$\frac{x}{2}$$ + C
= $$\frac{1}{2}$$ [x (1 + tan2 x) – tan x] + C
= $$\frac{1}{2}$$ [x sec2 x – tan x] + C

(ii) Let I = ∫ (elog x + sin x) cos x dx
= ∫ (x + sin x) cos x dx
= ∫ x cos x dx + ∫ $$\frac{1}{2}$$ sin 2x dx
= x sin x – ∫ 1 . sin x dx – $$\frac{\cos 2 x}{4}$$ + c
= x sin x + cos x – $$\frac{\cos 2 x}{4}$$ + c

Question 5.
(i) ∫ x log x dx (NCERT)
(ii) ∫ x log 2x (NCERT)
Solution:
(i) Let I = ∫ x log x dx
using integrating by parts
∴ I = log x . $$\frac{x^2}{2}$$ – ∫ $$\frac{1}{x} \cdot \frac{x^2}{2}$$ dx
= $$\frac{x^2}{2}$$ log x – $$\frac{x^2}{4}$$ + C
= $$\frac{x^2}{4}$$ (2 log x – 1) + C

(ii) ∫ x log 2x dx = log 2x . $$\frac{x^2}{2}$$ – ∫ $$\frac{2}{2 x} \frac{x^2}{2}$$ dx
[Integrating by parts]
= $$\frac{x^2}{2}$$ log 2x – $$\frac{x^2}{4}$$ + C
= $$\frac{x^2}{4}$$ [2 log 2x – 1] + C

Question 6.
(i) ∫ x4 log x dx
(ii) ∫ log x dx
Solution:
(i) Let I = ∫ x4 log x dx
= log x . $$\frac{x^5}{5}$$ – ∫ $$\frac{1}{x} \cdot \frac{x^5}{5}$$ dx + C
= $$\frac{x^5}{5} \log x-\frac{1}{5} \cdot \frac{x^5}{5}$$ + C
= $$\frac{x^5}{25}$$ (5 log x – 1) + C

(ii) ∫ log x dx = ∫ log x . 1 dx
= x log x – ∫ $$\frac{1}{x}$$ . x dx
= x log x – x + C
= x (log x – 1) + C

Question 6 (old).
(i) ∫ x2 log x dx (NCERT)
Solution:
(i) ∫ x2
= $$\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x$$
= $$\frac{x^3}{3} \log x-\frac{x^3}{9}$$ + C
= $$\frac{x^3}{9}$$ (3 log x – 1) + C

Question 7.
(i) ∫ (x2 + 1) log x dx (NCERT)
(ii) ∫ x2 log (1 + x) dx
Solution:
(i) ∫ (x2 + 1) log x dx
= log x . ($$\frac{x^3}{3}$$ + x) – ∫ $$\frac{1}{x}\left(\frac{x^3}{3}+x\right)$$
[using integrating by parts]
∴ I = ($$\frac{x^3}{3}$$ + x) log x – $$\frac{1}{3} \frac{x^3}{3}$$ – x + C
= $$\frac{x^3}{9}$$ (3 log x – 1) + x (log x – 1) + C

(ii) ∫ x2 log (1 + x) dx

Question 8.
(i) ∫ tan-1 x dx (NCERT)
(ii) ∫ cos-1 $$\left(\frac{1}{x}\right)$$ dx
Solution:
(i) ∫ tan-1 x dx
= ∫ tan-1 x . 1 dx
= x tan-1 x – ∫ $$\frac{x}{1+x^2}$$ dx
= x tan-1 x – $$\frac{1}{2}$$ log (1 + x2) + C
[∵ ∫ $$\frac{f^{\prime}(x)}{f(x)}$$ dx = log |f(x)| + C]

(ii) ∫ cos-1 $$\frac{1}{x}$$ dx
= ∫ sec-1 x dx
= ∫ sec-1 x . 1 dx
= x sec-1 x – ∫ $$\frac{1}{x \sqrt{x^2-1}}$$ x dx
= x sec-1 x – ∫ $$\frac{d x}{\sqrt{x^2-1}}$$
= x sec-1 x – log |x + $$\sqrt{x^2-1}$$ + C

Question 9.
(i) ∫ x3 tan-1 x dx
(ii) ∫ x2 sin-1 x dx
Solution:
(i) Let I = ∫ x3 tan-1 x dx

(ii) Let I = ∫ x2 sin-1 x dx

Question 10.
(i) ∫ $$\frac{\sin ^{-1} x}{\sqrt{1-x}}$$ dx
(ii) ∫ x2 ex dx
Solution:
(i) Let I = ∫ $$\frac{\sin ^{-1} x}{\sqrt{1-x}}$$ dx

(ii) ∫ x2 ex dx
= x2 ex – ∫ 2x . ex dx [Integrating by parts]
= x2 ex – 2 [x ex – ∫ 1 . ex dx] [Integrating by parts]
= x2 ex – 2 [x ex – ex] + C
= ex (x2 – 2x + 2) + C

Question 11.
(i) ∫ x2 e3 x dx
(ii) ∫ x2 cos x dx
Solution:
(i) ∫ x2 e3 x dx

(ii) ∫ x2 cos x dx
= x2 sin x – ∫ 2x sin x dx
= x2 sin x – 2 [- x cos x + ∫ 1 . cos x dx]
= x2 – 2 [- x cos x + ∫ 1 . cos x dx]
= x2 – 2 [- x cos x + sin x] + c
= x2 sin x + 2x cos x – 2 sin x + c

Question 12.
(i) ∫ (log x)2 dx
(ii) ∫ x3 sin (x2) dx
Solution:
(i) ∫ (log x)2 dx
= ∫ (log x)2 . 1 dx
= (log x)2 x – ∫ 2 log x . $$\frac{1}{x}$$ . x dx
= x (log x)2 – 2 ∫ log x . 1 dx
= x (log x)2 – 2 [x log x – x] + C
= x (log x)2 – 2x log x + 2x + C

(ii) Let I = ∫ x3 sin (x2) dx
put x2 = t
⇒ 2x dx = dt
= ∫ t sin t $$\frac{dt}{2}$$
= $$\frac{1}{2}$$ [- t cos t + ∫ 1 . cos t dt] + C
= $$\frac{1}{2}$$ [- t cos t + sin t] + C
= $$\frac{1}{2}$$ [- x2 cos2 x2 + sin x2] + C

Question 13.
(i) ∫ 2x3 ex2 dx
(ii) ∫ cos √x dx
Solution:
(i) Let I = ∫ 2x3 ex2 dx
put x2 = t
⇒ 2x dx = dt
= ∫ t . et dt
= t et – ∫ 1 . et dt + C
= t et – et + C
= (t – 1) et + C
= (x2 – 1) ex2 + C

(ii) Let I = ∫ cos √x + C
put √x = t
⇒ x = t2
⇒ dx = 2t dt
= ∫ cos t (2t dt)
= 2 ∫ t cos t dt
= 2 [t sin t – ∫ 1 . sin t dt] + c
= 2 [t sin t + cos t] + c
= 2 [√x sin √x + cos √x] + c

Question 14.
(i) ∫ tan-1 √x dx
(ii) ∫ x3 tan-1 (x2) dx
Solution:
(i) Let I = ∫ tan-1 √x dx
put √x = t
⇒ x = t2
⇒ dx = 2t dt
= ∫ tan-1 t (2t dt)
= 2 ∫ tan-1 t t dt
= 2 [tan-1 t . $$\frac{t^2}{2}$$ – ∫ $$\frac{1}{1+t^2} \cdot \frac{t^2}{2}$$] + c
= 2 $$\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{1+t^2-1}{1+t^2} d t\right]$$
= t2 tan-1 t – ∫ (1 – $$\frac{1}{1+t^2}$$) dt
= t2 tan-1 t – t + tan-1 t + c
= (1 + t2) tan-1 t – t + c
= (1 + x) tan-1 √x – √x + c

(ii) Let I = ∫ x3 tan-1 (x2) dx
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ t . tan-1 t $$\frac{dt}{2}$$

Question 15.
(i) ∫ sin3 √x dx
(ii) ∫ $$\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}$$ dx (NCERT)
Solution:
(i) Let I = ∫ sin3 √x dx
put √x = t
⇒ x = t2
⇒ dx = 2t dt
= ∫ sin3 t (2t dt)
= 2 ∫ $$\frac{t}{4}$$ [3 sin t – sin 3t] dt
[∵ sin 3t = 3 sin t – 4 sin3 t]
⇒ sin3 t = $$\frac{1}{4}$$ [3 sin t – sin 3t]

(ii) Let I = ∫ $$\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}$$ dx
put cos-1 x = t
x = cos t
dx = – sin t dt
= ∫ $$\frac{\cos t \cdot t(-\sin t d t)}{\sqrt{1-\cos ^2 t}}$$
= – ∫ t cos tdt
= – [t sin t – ∫ 1 . sin t dt] + c
= – [t sin t + cos t] + c
= – [$$\sqrt{1-x^2}$$ cos-1 x + x] + c

Question 16.
(i) ∫ sin-1 (3x – 4x3) dx
(ii) ∫ tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ dx
Solution:
(i) Let I = ∫ sin-1 (3x – 4x3) dx
put x = sin θ
⇒ dx = cos θ dθ
= ∫ sin-1 (3 sin θ – 4 sin3 θ) cos θ dθ
= ∫ sin-1 (sin 3θ) cos θ dθ
= 3 ∫ θ . cos θ dθ
= 3 [θ sin θ – ∫ 1 . sin θ] + c
= 3 [θ sin θ + cos θ] + c
= 3 [sin-1 x . x + $$\sqrt{1-x^2}$$] + c
[∵ cos θ = $$\sqrt{1-\sin ^2 \theta}$$
= $$\sqrt{1-x^2}$$]
= 3 [x sin-1 x + $$\sqrt{1-x^2}$$] + c

(ii) Let I = ∫ tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ dx
put x = tan θ
⇒ dx = sec2 θ dθ
= ∫ tan-1 $$\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$$ sec2 θ dθ
= ∫ tan-1 (tan 3θ) sec2 θ dθ
= 3 ∫ θ sec2 θ dθ

Question 17.
(i) ∫ $$\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$$ dx
(ii) ∫ $$\frac{x^2 \tan ^{-1} x}{1+x^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$$ dx
put x = sin θ
⇒ dx = cos θ dθ
∴ θ = sin-1 x
⇒ I = ∫ $$\frac{\theta \cos \theta d \theta}{\left(1-\sin ^2 \theta\right)^{3 / 2}}$$
= $$\frac{\theta \cos \theta d \theta}{\cos ^3 \theta}$$
= ∫ θ sec2 θ dθ
= θ tan θ – ∫ 1 . tan θ dθ
= θ tan θ – ∫ $$\frac{\sin \theta}{\cos \theta}$$ dθ
= θ tan θ + log |cos θ| + C

(ii) Let I = ∫ $$\frac{x^2 \tan ^{-1} x}{1+x^2}$$ dx
put tan-1 x = θ
⇒ x = tan θ
⇒ dx = sec2 θ dθ
= ∫ $$\frac{\tan ^2 \theta \cdot \theta}{1+\tan ^2 \theta}$$ . sec2 θ dθ
= ∫ θ tan2 θ dθ
= ∫ θ (sec2 θ – 1) dθ

= ∫ θ sec2 θ dθ – ∫ θ dθ
= θ tan θ – ∫ 1 . tan θ dθ – $$\frac{\theta^2}{2}$$ + c
= θ tan θ + log |cos θ| – $$\frac{\theta^2}{2}$$ + c
= x tan-1 x + log $$\left|\frac{1}{\sqrt{1+x^2}}\right|$$ – $$\frac{1}{2}$$ (tan-1 x)2 + c
= x tan-1 x – $$\frac{1}{2}$$ log |1 + x2| – $$\frac{1}{2}$$ (tan-1 x)2 + c

Question 18.
(i) ∫ ex sin x dx
(ii) ∫ e2x sin x dx
Solution:
(i) Let I = ∫ ex sin x dx
= sin x e x – ∫ cos x ex dx
= ex sin x – [cos x . ex – ∫ – sin x ex dx]
= (sin x – cos x) ex – 1
⇒ 2I = (sin x – cos x) ex – 1
⇒ I = $$\frac{e^x}{2}$$ (sin x – cos x) + C

[Students can also take first function as sin x]
= – e2x cos x + ∫ 2 e2x cos x dx
∴ I = – e2x cos x + 2 [e2x sin x – ∫ 2 e2x sin x dx]
⇒ I = e2x (- cos x + 2 sin x) – 4 I
⇒ I = $$\frac{e^{2 x}}{5}$$ [2 sin x – cos x] + C

Question 19.
(i) ∫ eax sin bx dx
(ii) ∫ eax cos (bx + c) dx
Solution:
(i) Let I = ∫ eax sin bx dx

(ii) Let I = ∫ eax cos (bx + c) dx
= cos (bx + c) $$\frac{e^{a x}}{a}$$ + ∫ sin (bx + c) . b . $$\frac{e^{a x}}{a}$$ dx

Question 20.
(i) ∫ x2 ex3 cos x3 dx
(ii) ∫ ex sin2 x dx
Solution:
(i) Let I = ∫ x2 ex3 cos x3 dx
put x3 = t
⇒ 3x2 dx = dt
= ∫ et cos t $$\frac{d t}{3}$$
= $$\frac{1}{3}$$ ∫ et cos t dt
= $$\frac{1}{3}$$ I1 ………………(1)
where I1 = ∫ et cos t dt
= et sin t – ∫ et sin t dt
= et sin t – [- et cos t – ∫ et (- cos t) dt]
∴ I1 = et sin t + et cos t – I1
⇒ I1 = $$\frac{e^t}{2}$$ [sin t + cos t]
∴ from (1) ;
∴ I = $$\frac{e^t}{6}$$ [sin t + cos t]
I = $$\frac{e^{x^3}}{6}$$ [sin x3 + cos x3] + c

(ii) Let I1 = ∫ ex sin2 x dx

Question 21.
(i) ∫ esin-1 x dx
(ii) ∫ $$\frac{e^{m \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}$$ dx
(iii) ∫ cosec3 x dx
Solution:
(i) put sin-1 x = t
⇒ x = sin t
⇒ dx = cos t dt
∴ I = ∫ esin-1 x dx
= ∫ et cos t dt
= et sin t – ∫ et sin t dt
∴ I = et sin t – [et (- cos t) + ∫ et cos t dt]
⇒ I = et (sin t + cos t) – I
⇒ I = $$\frac{e^t}{2}$$ (sin t + cos t) + C
⇒ I = $$\frac{e^{\sin ^{-1} x}}{2}\left[x+\sqrt{1-x^2}\right]$$ + C

(ii) Let I = ∫ $$\frac{e^m \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}}$$ dx
put tan-1 x = t
⇒ x = tan t
⇒ dx = sec2 t dt

(iii) Let I = ∫ cosec3 x dx
= ∫ cosec x cosec2 x dx
= cosec x (- cot x) – ∫ – cot x cosec x (- cot x) dx
= – cot x cosec x – ∫ cosec x (cosec2 x – 1) dx
= – cot x cosec x – ∫ cosec3 x dx + ∫ cosec x dx
⇒ I = – cot x cosec x – I + ∫ cosec x dx
⇒ 2I = – cot x cosec x + log |tan $$\frac{x}{2}$$| + c
⇒ I = $$\frac{-\cot x \ {cosec} x}{2}+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|$$ + c

Question 22.
(i) ∫ cos (log x) dx
(ii) ∫ $$\frac{\sin ^{-1} x}{x^2}$$ dx
Solution:
(i) Let I = ∫ cos (log x) . 1 dx
= cos (log x) . x – ∫ – sin (log x) . $$\frac{1}{x}$$ . x dx
= x cos (log x) + ∫ sin (log x) . 1 dx
= x cos (log x) + sin (log x) . x – ∫ cos (log x) . $$\frac{1}{x}$$ . x dx
∴ I = x [cos (log x) + sin (log x)] – I
⇒ 2I = x [cos (log x) + sin (log x)]
⇒ I = $$\frac{x}{2}$$ [cos (log x) + sin (log x)] + c

(ii) Let I = ∫ $$\frac{\sin ^{-1} x}{x^2}$$ dx
put sin-1 x = t
x = sin t
⇒ dx = cos t dt
= ∫ $$\frac{t}{\sin ^2 t}$$ cos t dt
= ∫ t (cot t cosec t dt)
= t (- cosec t) – ∫ 1 . (- cosec t) dt + c

= – t cosec t + ∫ cosec t dt + c
= – t cosec t – log |cosec t + cot t| + c
= $$-\frac{1}{x} \sin ^{-1} x-\log \left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|$$ + c
= $$\frac{-\sin ^{-1} x}{x}-\log \left|\frac{1+\sqrt{1-x^2}}{x}\right|$$ + c