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) ∫ (x2/3 + 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) ∫ (x2/3 + 1) dx
= ∫ x2/3 dx + ∫ 1 dx
= $$\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}$$ + x + C
= $$\frac{3}{5}$$ x5/3 + 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 ∫ x2/3 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 + $$\frac{b}{x}$$ + $$\frac{9 c}{5}$$ x5/3 + 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 + $$\frac{10}{3}$$ x3/2 + C

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

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

Question 5.
(i) ∫ (x3/2 + 2 ex – $$\frac{1}{x}$$) dx (NCERT)
(ii) ∫ (√x + $$\frac{1}{\sqrt{x}}$$) dx (NCERT)
Solution:
(i) ∫ (x3/2 + 2 ex – $$\frac{1}{x}$$) dx
= $$\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}$$ + 2 ex – log |x| + C
= $$\frac{2}{5}$$ x5/2 + 2 ex – 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
= $$\frac{2}{3}$$ x3/2 + 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 + ∫ x3/2 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
= ∫ 3x5/2 dx + ∫ 2x3/2 dx + ∫ 3x1/2 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
= $$\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}$$ + 2x3/2 + 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
= ∫ x5/2 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
= $$\frac{2}{7} x^{\frac{7}{2}}$$ + 2x3/2 + 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 + x3/2 + 3x3/2 (x + √x)] dx
= ∫ x3 dx + ∫ x3/2 dx + 3 ∫ x5/2 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
= ∫ x3/2 dx + a2 ∫ x– $$\frac{1}{2}$$ dx + 2a ∫ √x dx
= $$\frac{2}{5}$$ x5/2 + 2a2√x + $$\frac{4 a}{3}$$ x3/2 + 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
= ∫ sec2</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]
= $$\frac{2}{3}$$ x3/2 + $$\frac{1}{2}$$ 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.