ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.2

Continuous practice using ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.2 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.2

Very short answer type questions (1 to 14) :

Evaluate the following (1 to 20) integrals :

Question 1.
(i) ∫ (ax² + bx + c) dx (NCERT)
(ii) ∫ (x^2/3 + 1) dx. (NCFRT)
Solution:
(i) ∫ (ax² + bx + c) dx
= a ∫ x² dx + b ∫ x dx + ∫ c dx
= \(\frac{a x^3}{3}+\frac{b x^2}{2}\) + cx + C
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C; n ≠ 1]

(ii) ∫ (x^2/3 + 1) dx
= ∫ x^2/3 dx + ∫ 1 dx
= \(\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\) + x + C
= \(\frac{3}{5}\) x^5/3 + x + C

Question 2.
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx (NCERT Exampler)
Solution:
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ \(\sqrt{3x}\) dx + ∫ \(\frac{1}{\sqrt{x}}\)) dx
= √3 ∫ x^{\(\frac{1}{2}\)} dx + ∫ x^{\(-\frac{1}{2}\)} dx
= √3 \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{\sqrt{3}} x^{3 / 2}\) + 2√x + C

(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx
= ∫ \(\frac{2 a}{\sqrt{x}}\) dx – b ∫ x^{– 2} dx + 3c ∫ (x²)^1/3 dx
= 2a ∫ x^{\(-\frac{1}{2}\)} dx – b ∫ x^{– 2} dx + 3c ∫ x^2/3 dx
= 2a \(\frac{x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}-b \frac{x^{-2+1}}{(-2+1)}+3 c \frac{x^{\frac{2}{3}+1}}{\left(\frac{2}{3}+1\right)}\) + C
= 4a√x + \(\frac{b}{x}\) + \(\frac{9 c}{5}\) x^5/3 + C.

Question 3.
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
(ii) ∫ (2x² 3 sin x + 5√x) dx (NCERT)
Solution:
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
= \(\frac{2}{3}\) ∫ cot x cosec x dx
= – \(\frac{2}{3}\) cosec x + C

(ii) ∫ (2x² – 3 sin x + 5√x) dx
= 2 ∫ x² dx – 3 ∫ sin x dx + 5 ∫ √x dx
= 2 \(\frac{x^{3}}{3}\) – 3 (- cos x) + 5 \(\frac{x^{3 / 2}}{3 / 2}\) dx
= \(\frac{2}{3}\) x³ + 3 cos x + \(\frac{10}{3}\) x^3/2 + C

Question 4.
(i) ∫ (2x – 3 cos x + e^x) dx (NCERT)
(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx (NCERT)
Solution:
(i) ∫ (2x – 3 cos x + e^x) dx
= ∫ 2x dx + 3 ∫ cos x dx + ∫ e^x dx
= \(\frac{2 x^2}{2}\) – 3 sin x + e^x + C
= x² – 3 sin x + e^x + C

(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx
= ∫ x² dx – ∫ 1 dx
= \(\frac{x^{3}}{3}\) – x + C

Question 5.
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx (NCERT)
(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx (NCERT)
Solution:
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx
= \(\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + 2 e^x – log |x| + C
= \(\frac{2}{5}\) x^5/2 + 2 e^x – log |x| + C
[∵ ∫ \(\frac{d x}{x}\) = log |x| + C ;
∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) ; n ≠ – 1]

(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ √x dx + ∫ \(\frac{d x}{\sqrt{x}}\)
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{3}\) x^3/2 + 2 √x + C

Question 6.
(i) ∫ (1 – x) √x dx
(ii) ∫ √x (3x² + 2x + 3) dx (NCERT)
Solution:
(i) ∫ (1 – x) √x dx
= ∫ √x dx + ∫ x^3/2 dx
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}-\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}\) + C

(ii) ∫ √x (3x² + 2x + 3) dx
= ∫ 3x^5/2 dx + ∫ 2x^3/2 dx + ∫ 3x^1/2 dx
= \(\frac{3 x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{2 x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3 x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}\) + 2x^3/2 + C

Question 7.
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx (ISC 2018)
(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx
= ∫ \(\left[\frac{x^3}{x^2}+\frac{5 x^2}{x^2}+\frac{4 x}{x^2}+\frac{1}{x^2}\right]\) dx
= ∫ x dx + ∫ 5 dx + 4 ∫ \(\frac{d x}{x}\) + ∫ x^-2 dx
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| + \(\frac{x^{-2+1}}{-2+1}\) + C
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| – \(\frac{1}{x}\) + C

(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx
= ∫ x^5/2 dx + 3 ∫ √x dx + 4 ∫ x^{\(-\frac{1}{2}\)} dx
= \(\frac{x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^{3 / 2}}{\frac{3}{2}}+\frac{4 x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{7} x^{\frac{7}{2}}\) + 2x^3/2 + 8√x + C

Question 8.
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
(ii) ∫ (x + √x)³ dx
Solution:
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
= ∫ x² [4x² + \(\frac{1}{x^{2}}\) – 4] dx
= ∫ 4x⁴ dx + ∫ dx – 4 ∫ x² dx
= \(\frac{4 x^5}{5}+x-\frac{4 x^3}{3}\) + C

(ii) ∫ (x + √x)³ dx
= ∫ [x³ + x^3/2 + 3x^3/2 (x + √x)] dx
= ∫ x³ dx + ∫ x^3/2 dx + 3 ∫ x^5/2 dx + 3 ∫ x² dx
= \(\frac{x^4}{4}+\frac{2 x^{5 / 2}}{5}+\frac{3 x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^3}{3}\) + C
= \(\frac{x^4}{4}+\frac{2}{5} x^{5 / 2}+\frac{6}{7} x^{7 / 2}\) + x³ + C

Question 9.
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
Solution:
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
= ∫ \(\frac{x^2+a^2+2 a x}{\sqrt{x}}\) dx
= ∫ x^3/2 dx + a² ∫ x^{– \(\frac{1}{2}\)} dx + 2a ∫ √x dx
= \(\frac{2}{5}\) x^5/2 + 2a²√x + \(\frac{4 a}{3}\) x^3/2 + C

(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{(x-1)\left(x^2+1\right)}{(x-1)}\) dx
= ∫ (x² + 1) dx
= \(\frac{x^{3}}{3}\) + x + C

Question 10.
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx (NCERT)
(ii) ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx
= \(\int \frac{1}{\cos ^2 x} d x-\int \frac{\sin x}{\cos ^2 x} d x\)
= ∫ sec² x dx – ∫ tan x sec x dx
= tan x – sec x + C

(ii) Let I = ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx
= ∫ \(\left(\frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}\right)\) dx
= ∫ \(\frac{x^6-x^5}{x^4-x^3}\) dx
= = ∫ \(\frac{x^5(x-1)}{x^3(x-1)}\) dx
= \(\frac{x^{3}}{3}\) + C

Question 11.
(i) ∫ sec x (sec x + tan x) dx
(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
Solution:
(i) ∫ sec x (sec x + tan x) dx
= ∫ sec² x dx + ∫ sec x tan x dx
= tan x + sec x + C

(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
= ∫ \(\) × sin² x dx
= ∫ tan² x dx
= ∫ (sec² x – 1) dx
= tan x – x + C

Question 12.
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
Solution:
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
= ∫ cot x cosec x dx – ∫ tan x sec x dx – 3 ∫ (sec² x – 1) dx
= – cosec x – sec x – 3 tan x + 3x + C
[∵ ∫ cot x cosec x dx = – cosec x + C
and ∫ tan x sec x dx = sec x + C]

(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
= ∫ 2 sec² x dx – 3 ∫ tan x sec x dx
= 2 tan x – 3 sec x + C

Question 13.
(i) ∫ \(\frac{1}{1+\cos x}\) dx
(ii) ∫ \(\frac{1}{1+\sin x}\) dx
Solution:
(i) ∫ \(\frac{1}{1+\cos x}\) dx
= ∫ \(\frac{1}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}\)
= ∫ \(\frac{1-\cos x}{\sin ^2 x}\) dx
= ∫ cosec² dx – ∫ cot x cosec x dx
= – cot x + cosec x + C

(ii) ∫ \(\frac{1}{1+\sin x}\) dx
= ∫ \(\frac{(1+\sin x) d x}{1-\sin ^2 x}\)
= ∫ \(\frac{(1+\sin x)}{\cos ^2 x}\) dx
= ∫ sec^{2</sup x dx + ∫ tan x sec x dx
= tan x + sec x + C}

Question 14.
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx
(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx

(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{1-\cos ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx
= ∫ (1 – cos x) dx
= ∫ dx – ∫ cos x dx
= x – sin x + c

Question 15.
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
= ∫ √x dx – ∫ \(\frac{x}{2}\) cos \(\frac{x}{2}\) dx + ∫ 5 dx
= \(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) – ∫ \(\frac{1}{2}\) sin x dx + 5x
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{2}{3}\) x^3/2 + \(\frac{1}{2}\) cos x + 5x + C

(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x\)
= ∫ cosec² x dx – ∫ sec² x dx
= – cot x – tan x + C

Question 16.
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
Solution:
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫ \(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫ sec x (sec x – tan x) dx
[∵ sec² θ – tan² θ = 1]
= ∫ sec² x dx – ∫ sec x tan x dx + C
= tan x – sec x + C

(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
= ∫ \(\frac{\ {cosec} x(\ {cosec} x+\cot x) d x}{(\ {cosec} x-\cot x)(\ {cosec} x+\cot x)}\)
= ∫ \(\frac{\ {cosec}^2 x+\cot x \ {cosec} x d x}{\ {cosec}^2 x-\cot ^2 x}\)
[∵ 1 + cot² θ = cosec² θ ]
= ∫ cosec² x dx + ∫ cot x cosec x dx
= – cot x – cosec x + C

Question 17.
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
= ∫ \(\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^2+1}\) dx
= ∫ (x⁴ – x² + 1) dx
= ∫ x⁴ dx + ∫ x² dx + ∫ 1 dx
= \(\frac{x^5}{5}-\frac{x^3}{3}\) + x + c
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c ; n ≠ – 1]

(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= 6 ∫ \(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + 5 ∫ \(\frac{\cos ^3 x d x}{\cos ^2 x \sin ^2 x}\)
= 6 ∫ tan x sec x dx + 5 ∫ cot x cosec x dx
= 6 sec x – 5 cosec x + C

Question 18.
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
Solution:
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
= ∫ \(\sqrt{2 \cos ^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
= ∫ \(\sqrt{2 \sin ^2 x}\) dx
= √2 ∫ sin x dx
= – √2 cos x + C

Question 19.
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
Solution:
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
= ∫ cot^-1 \(\left(\frac{2 \sin x \cos x}{2 \sin ^2 x}\right)\) dx
= ∫ cot^-1 (cot x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
= ∫ tan^-1 \(\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}}\) dx
= ∫ tan^-1 (tan x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

Question 20.
(i) ∫ tan^-1 (cosec x – cot x) dx
(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
Solution:
(i) ∫ tan^-1 (cosec x – cot x) dx

(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
= ∫ cos^-1 (cos 2x) dx
[∵ cos 2θ = \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)]
= ∫ 2x dx
= x² + c

Question 21.
If f’(x) = 4x³ – 6 and f (0) = 3, find f (x).
Solution:
Given f’ (x) = 4x³ – 6
On integrating both sides w.r.t. x, we get
∫ f’(x) dx = ∫ (4x³ – 6) dx
f(x) = x⁴ – 6x + C ………..(1)
given f(0) = 3
i.e. when x = 0 ;
f(x) = 3
∴ from (1) ;
3 = c
Thus, from (1) ;
f(x) = x⁴ – 6x + 3.