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

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 e^x dx
(ii) ∫ x sin x dx (NCERT)
Solution:
(i) ∫ x e^x dx = x . e^x – ∫ 1 . e^x dx
[Integrating by parts]
= x e^x – e^x + C
= (x – 1) e^x + 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 e^2x dx
(ii) ∫ x cos 2x dx
Solution:
(i) ∫ x e^2x 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) e^2x + 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) ∫ (e^{log 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 = ∫ (e^{log 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 e^x dx
Solution:
(i) Let I = ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x}}\) dx

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

Question 11.
(i) ∫ x² e^{3 x} dx
(ii) ∫ x² cos x dx
Solution:
(i) ∫ x² e^{3 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 . e^t dt
= t e^t – ∫ 1 . e^t dt + C
= t e^t – e^t + C
= (t – 1) e^t + C
= (x² – 1) e^x2 + 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) ∫ e^x sin x dx
(ii) ∫ e^2x sin x dx
Solution:
(i) Let I = ∫ e^x sin x dx
= sin x e ^x – ∫ cos x e^x dx
= e^x sin x – [cos x . e^x – ∫ – sin x e^x dx]
= (sin x – cos x) e^x – 1
⇒ 2I = (sin x – cos x) e^x – 1
⇒ I = \(\frac{e^x}{2}\) (sin x – cos x) + C

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

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

(ii) Let I = ∫ e^ax 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 e^x3 cos x³ dx
(ii) ∫ e^x sin² x dx
Solution:
(i) Let I = ∫ x² e^x3 cos x³ dx
put x³ = t
⇒ 3x² dx = dt
= ∫ e^t cos t \(\frac{d t}{3}\)
= \(\frac{1}{3}\) ∫ e^t cos t dt
= \(\frac{1}{3}\) I₁ ………………(1)
where I₁ = ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
= e^t sin t – [- e^t cos t – ∫ e^t (- cos t) dt]
∴ I₁ = e^t sin t + e^t 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 = ∫ e^x sin² x dx

Question 21.
(i) ∫ e^{sin-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 = ∫ e^{sin-1 x} dx
= ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
∴ I = e^t sin t – [e^t (- cos t) + ∫ e^t cos t dt]
⇒ I = e^t (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