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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Very Short answer type questions (1 to 7):

Evaluate the following (1 to 4) :

Question 1.
If f'(x) = √x and f(1) = 2, then find the f(x).
Solution:
Since f(x) = ∫ f'(x) dx + C
⇒ f(x) = ∫ √x dx + C
[∵ f'(x) = √x]
⇒ f(x) = $$\frac{2}{3}$$ x3/2 + C ………..(1)
Since f(1) = 2
⇒ When x = 1 ; f(x) = 2
∴ from (1);
2 = $$\frac{2}{3}$$ + C
⇒ C = $$\frac{4}{3}$$
∴ f(x) = $$\frac{2}{3}$$ x3/2 + $$\frac{4}{3}$$

Question 1 (old).
(i) ∫ $$\frac{x}{\sqrt{1-x^2}}$$ dx
(ii) ∫ x2 ex3 dx
(iii) ∫ eex ex dx
Solution:
(i) Let I = ∫ $$\frac{x}{\sqrt{1-x^2}}$$ dx
put x2 = t
⇒ 2x dx = dt
= ∫ $$\frac{d t}{2 \sqrt{1-t}}$$
= $$\frac{1}{2}$$ ∫ (1 – t)– $$\frac{1}{2}$$
= $$\frac{1}{2} \frac{(1-t)^{-\frac{1}{2}+1}}{(-1)\left(-\frac{1}{2}+1\right)}$$ + C
= – $$\sqrt{1-x^2}$$ + C

(ii) Let I = ∫ x2 ex3 dx
put x3 = t
⇒ 3x2 dx = dt
= ∫ et $$\frac{d t}{3}$$
= $$\frac{e^{x^3}}{3}$$ + C

(iii) put ex = t
ex dx = dt
∴ I = ∫ eex ex dx
= ∫ et dt
= et + C
= eex + C

Question 2.
If f'(x) = 4x3 – $$\frac{3}{x^4}$$ and f(- 1) = 0, find f(x).
Solution:
Given f'(x) = 4x3 – $$\frac{3}{x^4}$$
On integrating both sides w.r.t x, we have
∫ f'(x) dx = ∫ (4x3 – $$\frac{3}{x^4}$$) dx + C
⇒ f(x) = $$\frac{4 x^4}{4}-3 \frac{x^{-4+1}}{-4+1}$$ + C
⇒ f(x) = x4 + $$\frac{1}{x^3}$$ + C …………….(1)
Given f(- 1) = 0 i.e.
When x = – 1 ;
f(x) = 0
0 = 1 – 1 + C
⇒ C = 0
∴ from (1) ;
f(x) = x4 + $$\frac{1}{x^3}$$

Question 2 (old).
(i) ∫ $$\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}$$
(ii) ∫ $$\sqrt{2+\sin 3 x}$$ cos 3x dx
(iii) ∫ $$\frac{d x}{4 \cos ^3 x-3 \cos x}$$
Solution:
(i) put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
∴ ∫ $$\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}$$ = ∫ $$\frac{d t}{t}$$
= log |t| + C
= log |tan-1 x| + C

(ii) Let I = ∫ $$\sqrt{2+\sin 3 x}$$ cos 3x dx
put sin 3x = t
⇒ 3 cos 3x dx = dt
∴ I = ∫ $$\sqrt{2+t} \frac{d t}{3}$$
= $$\frac{1}{3} \frac{(2+t)^{3 / 2}}{3 / 2}$$ + C
= $$\frac{2}{9}$$ (2 + sin 3x)3/2 + C

(iii) Let I = ∫ $$\frac{d x}{4 \cos ^3 x-3 \cos x}$$
= ∫ $$\frac{d x}{\cos 3 x}$$
= ∫ sec 3x dx
= $$\frac{1}{3}$$ log |sec 3x + tan 3x| + C

Question 3.
(i) ∫ sec x $$\sqrt{\frac{1-\sin x}{1+\sin x}}$$ dx
(ii) ∫ $$\frac{1}{\ {cosec} x-1}$$ dx
Solution:
(i) ∫ sec x $$\sqrt{\frac{1-\sin x}{1+\sin x}}$$ dx
= ∫ sec x $$\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}$$ dx
= ∫ $$\frac{\sec x(1-\sin x)}{\cos x}$$ dx
= ∫ $$\frac{(1-\sin x)}{\cos ^2 x}$$ dx
= ∫ sec2 x dx – ∫ tan x sec x dx + C
= tan x – sec x + C

(ii) ∫ $$\frac{1}{\ {cosec} x-1}$$ dx
= ∫ $$\frac{\sin x d x}{1-\sin x}$$ dx
= ∫ $$\frac{\sin x(1+\sin x)}{\cos ^2 x}$$ dx
= ∫ tan x sec x dx + ∫ (sec2 x – 1) dx
= sec x + tan x – x + C

Question 3 (old).
(i) $$\int_1^2 \frac{x}{x^2+1}$$ dx
(ii) $$\int_1^e \frac{1+\log x}{2 x}$$ dx
(iii) $$\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}$$ dx
Solution:
(i) Let I = $$\int_1^2 \frac{x}{x^2+1}$$ dx
put x2 + 1 = t
⇒ 2x dx = dt
When x = 1 ⇒ t = 2 ;
When x = 2 ⇒ t = 5
∴ I = $$\int_2^5 \frac{d t}{2 t}$$
= $$\left.\frac{1}{2} \log |t|\right]_2^5$$
= $$\frac{1}{2}$$ (log 5 – log 2)
= $$\frac{1}{2}$$ log $$\frac{5}{2}$$.

(ii) put 1 + log x = t
⇒ $$\frac{1}{x}$$ dx = dt
When x = 1
⇒ t = 1 + log 1 = 1
When x = e
⇒ t = 1 + log e = 1 + 1 = 2
∴ I = $$\int_1^2 \frac{1+\log x}{2 x}$$ dx
= $$\frac{1}{2} \int_1^2$$ t dt
= $$\left.\frac{t^2}{4}\right]_1^2$$
= $$\frac{1}{4}$$ [4 – 1]
= $$\frac{3}{4}$$

(iii) $$\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}$$ dx

Question 4.
(i) ∫ $$\frac{1+\sin ^2 x}{1+\cos x}$$ dx
(ii) ∫ (4 cot x – 5 tan x)2 dx
Solution:
(i) Let I = ∫ $$\frac{1+\sin ^2 x}{1+\cos x}$$ dx

(ii) Let I = ∫ (4 cot x – 5 tan x)2 dx
= ∫ (16 cot2 x + 25 tan2 x – 40) dx
= ∫ [16 (cosec2 x – 1) + 25 (sec2 x – 1) – 40] dx
= ∫ 16 cosec2 x dx + 25 ∫ sec2 x dx – 81 ∫ dx
= – cot x + 25 tan x – 81 x + C

Question 4 (old).
(ii) $$\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}$$ dx
Solution:
Let I = $$\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}$$ dx

Question 5.
(i) ∫ $$\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}$$
(ii) ∫ (1 – x) $$\sqrt{1+x}$$ dx
Solution:
(i) Let I = $$\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}$$

(ii) Let I = ∫ (1 – x) $$\sqrt{1+x}$$ dx
= ∫ – (x + 1 – 2) $$\sqrt{1+x}$$ dx
= – ∫ [(x + 1)3/2 – 2 (1 + x)1/2] dx
= – $$\frac{2}{5}$$ (x + 1)5/2 + $$\frac{4}{3}$$ (1 + x)3/2 + C

Question 5 (old).
If ∫ |x| dx = k x |x| + C, then what is the value of k ?
Solution:
Let I = ∫ |x| dx
= ∫ |x| . 1 dx
= |x| . x – ∫ $$\frac{x}{|x|}$$ . x dx
∴ I = x |x| – ∫ $$\frac{|x|^2}{|x|}$$ dx + 2C
⇒ I = x |x| – ∫ |x| dx + 2C
⇒ 2I = x |x| + 2C
⇒ I = $$\frac{x|x|}{2}$$ + C ………….(1)
Also I = kx |x| + C …………….(2)
∴ from (1) and (2) ;
we have k = $$\frac{1}{2}$$.

Question 6.
(i) ∫ (x + 1) (2x – 1)3/2 dx
(ii) ∫ sec2 2x cos 4x dx
Solution:
(i) Let I = ∫ (x + 1) (2x – 1)3/2 dx
= $$\frac{1}{2}$$ ∫ (2x – 1 + 3) (2x – 1)3/2 dx
= $$\frac{1}{2} \int(2 x-1)^{5 / 2} d x+\frac{3}{2} \int(2 x-1)^{3 / 2}$$ dx
= $$\frac{1}{2} \frac{(2 x-1)^{7 / 2}}{\frac{7}{2} \times 2}+\frac{3}{2} \frac{(2 x-1)^{5 / 2}}{\frac{5}{2} \times 2}$$ + C
= $$\frac{1}{14}$$ (2x – 1)3/2 + $$\frac{3}{10}$$ (2x – 1)5/2 + C

(ii) ∫ sec2 2x cos 4x dx
= ∫ sec2 2x cos (2 × 2x) dx
= ∫ sec2 2x (2 cos2 2x – 1) dx
= ∫ [2 – sec2 2x] dx
= 2x – $$\frac{\tan 2 x}{2}$$ + C

Question 7.
(i) ∫ $$\frac{10 x^9+10^x \log 10}{x^{10}+10^x}$$ dx (NCERT)
(ii) ∫ $$\frac{d x}{x \sqrt{a x-x^2}}$$ dx (NCERT)
Solution:
(i) put x10 + 10x = t
⇒ (10x9 + 10x log 10) dx = dt
∴ ∫ $$\frac{10 x^9+10^x \log 10}{x^{10}+10^x}$$ dx = ∫ $$\frac{d t}{t}$$
= log |t| + C
= log |x10 + 10x| + C

(ii) Let I = ∫ $$\frac{d x}{x \sqrt{a x-x^2}}$$ dx

Question 8.
(i) ∫ $$\frac{d x}{\cos (x+a) \cos (x+b)}$$
(ii) ∫ $$\frac{d x}{3 \cos x+4 \sin x}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{\cos (x+a) \cos (x+b)}$$

(ii) Let I = ∫ $$\frac{d x}{3 \cos x+4 \sin x}$$
put 3 = r cos α …………..(1)
and 4 = r sin α …………..(2)
On squaring (1) and (2) ; we have
r = $$\sqrt{9+16}$$ = 5
On dividing (2) by (1) ; we have
tan α = $$\frac{4}{3}$$
⇒ α = tan-1 $$\frac{4}{3}$$
∴ I = ∫ $$\frac{d x}{r(\cos \alpha \cos x+\sin \alpha \sin x)}$$
= ∫ $$\frac{d x}{5 \cos (x-\alpha)}$$
= $$\frac{1}{5}$$ log |sec (x – α) + tan (x – α)| + C
where α = tan-1 $$\frac{4}{3}$$

Question 9.
(i) ∫ $$\frac{\sin 3 x}{\sin x}$$ dx
(ii) ∫ $$\frac{\sec x}{\sec 2 x}$$ dx
Solution:
(i) ∫ $$\frac{\sin 3 x}{\sin x}$$ dx
= ∫ $$\left[\frac{3 \sin x-4 \sin ^3 x}{\sin x}\right]$$ dx
= ∫ [3 – 4 sin2 x] dx
= ∫ [3 – 4 $$\left(\frac{1-\cos 2 x}{2}\right)$$] dx
= $$\frac{1}{2}$$ ∫ [2 + 4 cos 2x] dx
= ∫ (1 + 2 cos 2x) dx
= x + $$\frac{2 \sin 2 x}{2}$$ + C
= x + sin 2x + C

(ii) ∫ $$\frac{\sec x}{\sec 2 x}$$ dx
= ∫ $$\frac{\cos 2 x}{\cos x}$$ dx
= ∫ $$\left(\frac{2 \cos ^2 x-1}{\cos x}\right)$$ dx
= 2 ∫ cos x dx – ∫ sec x dx
= 2 sin x – log |sec x + tan x| + C

Question 10.
(i) ∫ $$\frac{e^x}{\sqrt{e^{2 x}-4}}$$ dx
(ii) ∫ ecot x cosec2 x dx
Solution:
(i) put ex = t
⇒ ex dx = dt
∴ I = ∫ $$\frac{e^x}{\sqrt{e^{2 x}-4}}$$ dx
= ∫ $$\frac{d t}{\sqrt{t^2-2^2}}$$
= log |t + $$\sqrt{t^2-2^2}$$| + C
= log |ex + $$\sqrt{e^{2 x}-4}$$| + C

(ii) Let I = ∫ ecot x cosec2 x dx
put cos x = t
⇒ – cosec2 x dx = dt
∴ I = ∫ et (- dt)
= – et + C
= – ecot x + C

Question 11.
(i) ∫ $$\frac{d x}{\sqrt{\sin ^3 x \cos x}}$$
(ii) ∫ $$\frac{\sin 2 x}{(a+b \cos 2 x)^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{\sin ^3 x \cos x}}$$

(ii) Let I = ∫ $$\frac{\sin 2 x}{(a+b \cos 2 x)^2}$$ dx
put cos 2x = t
⇒ – 2 sin 2x dx = dt
= ∫ $$\frac{d t}{-2(a+b t)^2}$$
= – $$\frac{1}{2} \frac{(a+b t)^{-2+1}}{(-2+1) b}$$ + C
= $$\frac{1}{2 b} \frac{1}{(a+b \cos 2 x)}$$ + C

Question 12.
(i) ∫ $$\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}$$ dx
(ii) ∫ $$\frac{x^3}{\sqrt{1+x^2}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}$$ dx

(ii) put x2 = t
⇒ 2x dx = dt

Question 13.
(i) ∫ $$\frac{\sec x}{\log (\sec x+\tan x)}$$ dx
(ii) ∫ $$\frac{\cos x-\sin x}{1-\sin 2 x}$$ dx
Solution:
(i) put log (sec x + tan x) = t
⇒ $$\frac{1}{\sec x+\tan x}$$ (sec x tan x + sec2 x) dx = dt
⇒ $$\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}$$ dx = dt
⇒ sec x dx = dt
∴ I = ∫ $$\frac{\sec x d x}{\log (\sec x+\tan x)}$$
= ∫ $$\frac{d t}{t}$$
= log |t| + C

(ii) Let I = ∫ $$\frac{\cos x-\sin x}{1-\sin 2 x}$$ dx

Question 14.
(i) ∫ $$\sqrt{1+2 \tan x(\tan x+\sec x)}$$ dx
(ii) ∫ $$\frac{4 x+1}{\sqrt{2 x^2+x-3}}$$ dx
Solution:
(i) Let I = ∫ $$\sqrt{1+2 \tan x(\tan x+\sec x)}$$ dx

(ii) put 2x2 + x – 3 = t
⇒ (4x + 1) dx = dt
∴ I = ∫ $$\frac{(4 x+1) d x}{\sqrt{2 x^2+x-3}}$$
= ∫ $$\frac{d t}{\sqrt{t}}$$
= ∫ t-1/2 dt
= $$\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$ + C
= 2 $$\sqrt{2 x^2+x-3}$$ + C

Question 15.
∫ $$\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}$$ dx, a ≠ ± b. What happens if a = ± b ?
Solution:
Case – I: When a ≠ ± b
Let I = ∫ $$\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}$$ dx
put a2 sin2 x + b2 cos2 x = t
⇒ [2a2 sin x cos x + 2b2 cos x (- sin x)] dx = dt
⇒ (a2 – b2) sin 2x dx = dt

Question 16.
(i) ∫ $$\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}$$
(ii) ∫ $$\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}$$

(ii) Let I = ∫ $$\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}$$ dx
put $$\sqrt{x^2+3}$$ = t
⇒ $$\frac{1}{2}\left(x^2+3\right)^{-\frac{1}{2}}$$ 2x dx = dt
⇒ $$\frac{x}{\sqrt{x^2+3}}$$ dx = dt
⇒ I = ∫ et dt
= et + C
Thus, I = $$e^{\sqrt{x^2+3}}$$ + C

Question 17.
(i) ∫ tan x sec2 x $$\sqrt{1-\tan ^2 x}$$ dx
(ii) ∫ sin3 x cos5 x dx
Solution:
(i) Let I = ∫ tan x sec2 x $$\sqrt{1-\tan ^2 x}$$ dx
put tan2 x dx = t
⇒ 2 tan x sec2 x dx = dt
∴ I = $$\frac{1}{2}$$ ∫ $$\sqrt{1-t}$$ dt
= $$\frac{1}{2} \frac{(1-t)^{3 / 2}}{\frac{3}{2}(-1)}$$ + C
= – $$\frac{1}{3}$$ (1 – tan2 x)3/2 + C

(ii) Let I = ∫ sin3 x cos5 x dx
= ∫ sin2 x cos5 x sin x dx
= ∫ (1 – cos2 x) cos5 x (sin x) dx
put cos x = t
⇒ – sin x dx = dt
= ∫ (1 – t2) t5 (- dt)
= – $$\left[\frac{t^6}{6}-\frac{t^8}{8}\right]$$ + C
= $$-\frac{\cos ^6 x}{6}+\frac{\cos ^8 x}{8}$$ + C

Question 18.
(i) ∫ $$\frac{(a+\sqrt{x})^n}{\sqrt{x}}$$ dx, n ≠ – 1
(ii) ∫ $$\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}$$
Solution:
(i) Let I = ∫ $$\frac{(a+\sqrt{x})^n}{\sqrt{x}}$$ dx, n ≠ – 1
put √x = t
⇒ $$\frac{1}{2 \sqrt{x}}$$ dx = dt
∴ I = ∫ (a + t)n (2 dt)
= $$\frac{2(a+t)^{n+1}}{n+1}$$ + C
= $$\frac{2}{n+1}$$ (a + √x)n + 1 + C ; n ≠ – 1

(ii) Let I = ∫ $$\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}$$

Question 19.
(i) ∫ $$\frac{d x}{(2 \sin x+3 \cos x)^2}$$
(ii) ∫ $$\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{(2 \sin x+3 \cos x)^2}$$
Divide Numerator and deno. by cos2 x ; we have
= ∫ $$\frac{\sec ^2 x d x}{(2 \tan x+3)^2}$$
put tan x = t
⇒ sec2 x dx = dt
∴ I = ∫ $$\frac{d t}{(2 t+3)^2}$$
= ∫ (2t + 3)-2 dt
= $$\frac{(2 t+3)^{-2+1}}{2 \cdot(-2+1)}$$ + C
= $$\frac{1}{2(2 t+3)}$$ + C
= – $$\frac{1}{2(2 \tan x+3)}$$ + C

(ii) Let I = ∫ $$\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}$$

Question 20.
(i) ∫ x $$\sqrt[3]{2 x+1}$$ dx
(ii) ∫ $$\frac{2 x}{x^2+3 x+2}$$ dx (NCERT)
Solution:
(i) Let I = ∫ x $$\sqrt[3]{2 x+1}$$ dx
put $$\sqrt[3]{2 x+1}$$ = t
⇒ 2x + 1 = t3
⇒ 2 dx = 3t2 dt

(ii) Let I = ∫ $$\frac{2 x}{x^2+3 x+2}$$ dx

Question 21.
(i) ∫ $$\frac{\sqrt{x}}{\sqrt{x}+2}$$ dx
(ii) ∫ √x (log x)2 dx
Solution:
(i) put √x + 2 = t
⇒ √x = t – 2
⇒ x = (t – 2)2
⇒ dx = 2 (t – 2) dt

= t2 – 8t + 8 log |t| + C
= (√x + 2)2 – 8 (√x + 2) + 8 log |t| + C
= x – 4√x + 8 log |√x + 2| + C’

(ii) Let I = ∫ √x (log x)2 dx
put log x = t
⇒ x = et
⇒ dx = et dt
∴ I = ∫ et/2 t2 et dt

Question 22.
(i) ∫ $$\frac{1}{x-x^3}$$ dx (NCERT)
(ii) ∫ $$\frac{\cos x}{2+\cos ^2 x}$$ dx
Solution:
(i) Let I = ∫ $$\frac{1}{x-x^3}$$ dx

(ii) Let I = ∫ $$\frac{\cos x}{2+\cos ^2 x}$$ dx

Question 23.
(i) ∫ $$\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}$$ dx (NCERT Exemplar)
(ii) ∫ $$\frac{e^x}{e^{2 x}+6 e^x+5}$$ dx
Solution:
(i) put x2 = y

(ii) Let I = ∫ $$\frac{e^x}{e^{2 x}+6 e^x+5}$$ dx ;
put ex = t
⇒ ex dx = dt

Question 24.
(i) ∫ $$\frac{\sin x}{(2+\cos x)(5+\cos x)}$$ dx
(ii) ∫ $$\frac{x^4}{(x-1)\left(x^2+1\right)}$$ dx
Solution:
(i) I = ∫ $$\frac{\sin x}{(2+\cos x)(5+\cos x)}$$ dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ $$\frac{x^4}{(x-1)\left(x^2+1\right)}$$ dx

Question 25
(i) ∫ cos 2x log (1 + tan x) dx
(ii) ∫ $$\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}$$ dx
Solution:
(i) Let I = ∫ cos 2x log (1 + tan x) dx

sin x = l (sin x + cos x) + m $$\frac{d}{d x}$$ (sin x + cos x)
⇒ sin x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x ;
1 = l – m ;
Coeff. of cos x ;
0 = l + m
On solving these eqn.’s ;
l = $$\frac{1}{2}$$ ; m = – $$\frac{1}{2}$$

(ii) Let I = ∫ $$\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}$$ dx

Question 25 (old).
(ii) ∫ $$\frac{1-\cos x}{\cos x(1+\cos x)}$$ dx
Solution:
Let I = ∫ $$\frac{1-\cos x}{\cos x(1+\cos x)}$$ dx

Question 26.
(i) ∫ $$\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}$$ dx
(ii) ∫ sin4 x cos5 x dx
Solution:
(i) Let I = ∫ $$\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}$$ dx ;
put x4 = t
⇒ 4x3 dx = dt

(ii) Let I = ∫ sin4 x cos5 x dx
= ∫ sin4 x cos4 x cos x dx
= ∫ sin4 x (1 – sin2 x)4 cos x dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ t4 (1 – t2)2 dt
= ∫ t4 (t4 – 2t2 + 1) dt
= $$\frac{t^9}{9}-2 \frac{t^7}{7}+\frac{t^5}{5}$$ + C
= $$\frac{\sin ^9 x}{9}-\frac{2}{7} \sin ^7 x+\frac{1}{5} \sin ^5 x$$ + C

Question 27.
(i) ∫ $$\frac{d x}{x^{1 / 2}-x^{1 / 4}}$$
(ii) ∫ tan-1 $$\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$ dx
Solution:
(i) Let I = $$\frac{d x}{x^{1 / 2}-x^{1 / 4}}$$
put x1/4 = t
⇒ x = t4
⇒ dx = 4t3 dt

= 2t2 + 4t + 4 log |t – 1| + C
= 2√x + 4x1/4 + 4 log |x1/4 – 1| + C

(ii) Let I = ∫ tan-1 $$\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$ dx
put x = cos 2t
⇒ dx = – 2 sin 2t dt

Question 28.
(i) ∫ $$\frac{\log (x+1)-\log x}{x(x+1)}$$ dx
(ii) ∫ $$\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}$$ ex dx
Solution:
(i) Let I = ∫ $$\frac{\log (x+1)-\log x}{x(x+1)}$$ dx ;
put log (x + 1) – log x = t

(ii) Let I = ∫ $$\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}$$ ex dx

Question 29.
(i) ∫ x3 e– x2 dx
(ii) ∫ sin-1 $$\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)$$ dx
Solution:
(i) Let I = ∫ x3 e– x2 dx ;
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ t e-t $$\frac{d t}{2}$$
= $$\frac{1}{2}\left[t \frac{e^{-t}}{(-1)}-\int 1 \times \frac{e^{-t}}{(-1)} d t\right]$$ + C
∴ I = $$\frac{1}{2}$$ [- t e-t – e-t] + C
= – $$\frac{1}{2}$$ (t + 1) e-t + C
= – $$\frac{1}{2}$$ (x2 + 1) e– x2 + C

(ii) Let I = ∫ sin-1 $$\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)$$ dx
first of all, we convert sin-1 to tan-1 with p = 2x + 2;
h = $$\sqrt{4 x^2+8 x+3}$$

Question 30.
(i) ∫ f'(ax + b) (f (ax + b))n dx (NCERT)
(ii) ∫ $$\frac{\sqrt{1-x}}{x^{7 / 2}}$$ dx
Solution:
(i) Case – I :
When n ≠ – 1
Let I = ∫ f'(ax + b) (f (ax + b))n dx
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
∴ I = ∫ $$\frac{1}{a}$$ tn dt
= $$\frac{1}{a} \frac{t^{n+1}}{n+1}$$ + C ; n ≠ – 1
∴ I = $$\frac{(f(a x+b))^{n+1}}{(n+1) a}$$ + C ; n ≠ – 1

Case – II :
When n = – 1
I = ∫ $$\frac{f^{\prime}(a x+b) d x}{f(a x+b)}$$
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
= ∫ $$\frac{d t}{a t}$$
= $$\frac{1}{a}$$ log |t| + C
= $$\frac{1}{a}$$ log |f (ax + b)| + C

(ii) Let I = ∫ $$\frac{\sqrt{1-x}}{x^{7 / 2}}$$ dx

Question 31.
(i) ∫ $$\sqrt{4-x+x^2}$$ dx
(ii) ∫ $$\frac{\tan ^{-1} x}{x^2}$$ dx
(iii) ∫ $$\frac{\log |x|}{(x+1)^3}$$ dx
Solution:
(i) Let I = ∫ $$\sqrt{4-x+x^2}$$ dx

(ii) Let I = ∫ $$\frac{\tan ^{-1} x}{x^2}$$ dx
put tan-1 x = θ
⇒ x = tan θ
⇒ dx = sec2 θ
= ∫ $$\frac{\theta}{\tan ^2 \theta}$$ sec2 θ dθ
= ∫ θ cosec2 θ dθ
= θ (- cot θ) – ∫ (- cot θ) dθ + C
= – θ cot θ + log |sin θ| + C
= $$-\frac{\tan ^{-1} x}{x}+\log \left|\frac{x}{\sqrt{1+x^2}}\right|$$ + C

(iii) Let I = ∫ $$\frac{\log |x|}{(x+1)^3}$$ dx
= ∫ log |x| . $$\frac{1}{(x+1)^3}$$ dx

Multiplying both sides of eqn. (2) by x(x + 1)2; we get
1 = A(x + 1)2 + Bx (x + 1) + Cx ………….(3)
putting x = 0, – 1 successively in eqn. (3); we have
1 = A
and 1= – C
⇒ C = – 1
Coeff. of x2;
0 = A + B,
⇒ B = – 1
∴ I1 = ∫ $$\left[\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}\right]$$ dx
= log |x| – log |x + 1| + $$\frac{1}{x+1}$$
∴ from (1); we have
= $$-\frac{\log |x|}{2(x+1)^2}+\frac{1}{2} \log \left|\frac{x}{x+1}\right|+\frac{1}{2(x+1)}$$ + C

Question 32.
Evaluate the following definite integrals as the limit of sums :
(i) $$\int_1^3$$ (x2 + 5x) dx
(ii) $$\int_0^1$$ e2 – 3x dx (NCERT)
Solution:
(i) On comparing $$\int_1^3$$ (x2 + 5x) dx with $$\int_a^b$$ f(x) dx
Here f(x) = x2 + 5x ;
a = 1 ; b = 3
∴ nh = b – a
= 3 – 1 = 2
∴ f(a) = f(1)
= 12 + 5
f(a + h) = f(1 + h)
= (1 + h)2 + 5 (1 + h)
f(a + 2h) = f(1 + 2h)
= (1 + 2h)2 + 5 (1 + 2h)
……………………………………
……………………………………

(ii) On comparing$$\int_0^1$$ e2 – 3x dx with $$\int_a^b$$ f(x) dx
Here, f(x) = e2-3x;
a = 0;
b = 1;
nh = b – a = 1 – 0 = 1
∴ f(a) = f(0) = e2
f(a + h) = f(h) = e2 – 3h
f(a + 2h) = f(2h) = e2 – 6h
……………………………
……………………………
$$f(a+\overline{n-1} h)$$ = f((n – 1)h)
= e2 – 3 (n – 1) h
∴ $$\int_0^1$$ e2 – 3h dx = $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h[f(a) + f(a + h) + f(a + 2h) ………..+ $$f(a+\overline{n-1} h)$$]
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h [e2 + e2 – 3h + e2 – 6h + ……. + e2 – 3(n – 1) h]
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h[1 + e– 3h + e– 6h ……… n terms]

Evaluate the following (33 to 36) definite integrals:

Question 33.
(i) $$\int_0^{\pi / 4}$$ tan2 x dx
(ii) $$\int_0^2 \frac{5 x+1}{x^2+4}$$ dx
Solution:
(i) $$\int_0^{\pi / 4}$$ tan2 x dx
= $$\int_0^{\pi / 4}$$ (sec2 – 1) dx
= tan x – x$$]_0^{\pi / 4}$$
= (tan $$\frac{\pi}{4}$$ – $$\frac{\pi}{4}$$) – (0 – 0)
= 1 – $$\frac{\pi}{4}$$

(ii) $$\int_0^2 \frac{5 x+1}{x^2+4}$$ dx

Question 34.
(i) $$\int_0^{\pi / 6}$$ (2 + 3x2) cos 3x dx
(ii) $$\int_0^1$$ (cos-1 x)2 dx
Solution:
(i) $$\int_0^{\pi / 6}$$ (2 + 3x2) cos 3x dx

(ii) Let I = $$\int_0^1$$ (cos-1 x)2 dx
put cos-1 x = t
⇒ – $$\frac{1}{\sqrt{1-x^2}}$$ dx = dt
⇒ x = cos t
⇒ dx = – sin t dt
When x = 0
⇒ t = $$\frac{\pi}{2}$$ ;
When x = 1
⇒ t = 0
∴ I = $$\int_{\pi / 2}^0$$ t2 (- sin t) dt
= – [- t2 cos t$$\}_{\pi / 2}^0$$ + $$\int_{\pi / 2}^0$$ t cos t dt]
∴ I = (0 × 1 – $$\frac{\pi^2}{4}$$ × 0) – 2 [t sin t$$\}_{\pi / 2}^0$$ – $$\int_{\pi / 2}^0$$ sin t dt]
∴ I = – 2 [t sin t + cos t$$]_{\pi / 2}^0$$
= + 2 [t sin t + cos t$$t]_0^{\pi / 2}$$
= + 2 [$$\frac{\pi}{2}$$ × 1 + 0 – 0 – 1]
= + 2 ($$\frac{\pi}{2}$$ – 1)
= π – 2

Question 35.
(i) $$\int_0^\pi$$ x sin x cos2 x dx (NCERT Exemplar)
(ii) $$\int_0^{\pi / 2}$$ $$\sqrt{sin x}$$ cos5 x dx (NCERT)
Solution:
(i) Let I = $$\int_0^\pi$$ x sin x cos2 x dx ……………(1)
∴ I = $$\int_0^\pi$$ (π – x) sin (π – x) cos2 (π – x) dx
I = $$\int_0^\pi$$ (π – x) sin x cos2 x dx ……………(2)
[∵ $$\int_0^a$$ f(x) dx = $$\int_0^a$$ f(a – x) dx]
On adding (1) and (2) ; we have
2I = $$\int_0^\pi$$ π sin x cos2 x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = π ⇒ t = – 1
⇒ 2I = π $$\int_1^{-1}$$ t2 (- dt)
= – π $$\left.\frac{t^3}{3}\right]_1^{-1}$$
= – $$\frac{\pi}{3}$$ (- 1 – 1)
= $$\frac{2 \pi}{3}$$
Thus, I = $$\frac{\pi}{3}$$

(ii) Let I = $$\int_0^{\pi / 2}$$ $$\sqrt{sin x}$$ cos5 x dx
= $$\int_0^{\pi / 2}$$ $$\sqrt{sin x}$$ (1 – sin2 x)2 cos x dx
put sin x = t ⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = $$\frac{\pi}{2}$$ ⇒ t = 1
∴ I = $$\int_0^1$$ √t (1 – t2)2 dt
= $$\int_0^1$$ √t (t4 – 2t2 + 1) dt
= $$\left.\frac{2 t^{11 / 2}}{11}-\frac{2 t^{7 / 2}}{7 / 2}+\frac{t^{3 / 2}}{3 / 2}\right]_0^1$$
= $$\left[\frac{2}{11}-\frac{4}{7}+\frac{2}{3}\right]$$
= $$\frac{42-132+154}{231}=\frac{64}{231}$$.

Question 36.
(i) $$\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}$$ dx
(ii) $$\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}$$ dx
Solution:
(i) I = $$\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}$$ dx

(ii) Let I = $$\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}$$ dx

put sin x – cos x = t
⇒ (cos x + sin x) dx = dt
On squaring; we have
(sin x – cos x)2 = t2
⇒ sin2 x + cos2 x – sin 2x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2
When x = 0 ⇒ t = – 1 ;
When x = $$\frac{\pi}{4}$$ ⇒ t = 0
∴ I = 4 $$\int_{-1}^0 \frac{d t}{4-\left(1-t^2\right)^2}$$ ………………(1)
= 4 $$\int_{-1}^0 \frac{d t}{\left(2+1-t^2\right)\left(2-1+t^2\right)}$$
⇒ I = 4 $$\int_{-1}^0 \frac{d t}{\left(3-t^2\right)\left(1+t^2\right)}$$
put t2 = y
Then $$\frac{1}{\left(3-t^2\right)\left(1+t^2\right)}=\frac{1}{(3-y)(1+y)}=\frac{\mathrm{A}}{3-y}+\frac{\mathrm{B}}{1+y}$$ …………….(2)
Multiply both sides of eqn. (2) by (3 – y) (1 + y) ; we get
1 = A (1 + y) + B (3 – y) ………….(3)
putting y = – 1, 3 successively in eqn. (3) ; we have
1 = 4B
⇒ B = $$\frac{1}{4}$$
and 1 = 4A
⇒ A = $$\frac{1}{4}$$
∴ from (2) ; we get ;

By using properties of definite integrals, evaluate the following (37 to 39) :

Question 37.
$$\int_0^4$$ (|x| + |x – 2| + |x – 4|
Solution:
Let f(x) = |x| + |x – 2| + |x – 4|
The critical points of f(x) are given by x = 0,2, 4
When 0 ≤ x < 2
⇒ |x| = x ;
x – 2 < 0
⇒ |x – 2| = – (x – 2)
and x – 4< 0
⇒ |x – 4| = – (x – 4)
∴ f(x) = x – (x – 2) – (x – 4) = – x + 6
When 2 ≤ x < 4
Then x – 2 ≥ 0;
x – 4 < 0; x > 0
∴ f(x) = x + x – 2 – (x – 4) = x+ 2
∴ $$\int_0^4$$ f(x) dx = $$\int_0^2$$ f(x) dx + $$\int_2^4$$ f(x) dx
= $$\int_0^2$$ (- x + 6) dx + $$\int_2^4$$ (x + 2) dx
= $$\left.\left[-\frac{x^2}{2}+6 x\right]_0^2+\frac{(x+2)^2}{2}\right]_2^4$$
= [(- 2 + 12) – (0 + 0)] + $$\frac{1}{2}$$ [36 – 16]
= 10 + 10 = 20.

Question 38.
(i) $$\int_{1 / e}^e$$ |log x| dx
(ii) $$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$$
(iii) $$\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}$$ dx
Solution:
(i) When $$\frac{1}{e}$$ ≤ x ≤ 1 ;
log x ≤ 0
∴ |log x| = – log x
When 1 ≤ x ≤ e ;
log x ≥ 0
∴ |log x| = + log x

(ii) Let I = $$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$$ …………….(1)

(iii) Let I = $$\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}$$ dx

Question 39.
$$\int_0^1$$ x (1 – x)5 dx
Solution:
Let I = $$\int_0^1$$ x (1 – x)5 dx
∴ I = $$\int_0^1$$ (1 – x) (1 – (1 – x))5 dx
[∵ $$\int_0^a$$ f(x) dx = $$\int_0^a$$ f (a – x) dx]
I = $$\int_0^1$$ (1 – x) x5 dx
= $$\left[\frac{x^6}{6}-\frac{x^7}{7}\right]_0^1$$
= $$\left(\frac{1}{6}-\frac{1}{7}-0-0\right)$$
= $$\frac{1}{42}$$

(ii) Let I = $$\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}$$ dx

Question 40.
(i) $$\int_{-\pi / 2}^{\pi / 2}$$ (x3 + x cos x + tan5 x + 1) dx
(ii) $$\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}$$ dx
Solution:
(i) Here f(x) = x3 + x cos x + tan5 x + 1
∴ f (- x) = (- x)3 + (- x) cos (- x) + [tan (- x)5] + 1
= – x3 – x cos x – tan5 x + 1
So f(- x) is neither equal to f(x) nor equal to – f(x)
Let I = $$\int_{-\pi / 2}^{\pi / 2}$$ [x3 + x cos x + tan5 x + 1] dx
= $$\int_{-\pi / 2}^{\pi / 2} x^3 d x+\int_{-\pi / 2}^{\pi / 2} x \cos x+\int_{-\pi / 2}^{\pi / 2} \tan ^5 x+\int_{-\pi / 2}^{\pi / 2} 1 d x$$ …………..(1)
Let f(x) = x3 ;
f(- x) = (- x)3
= – x3
= – f(x)
∴ $$\int_{-\pi / 2}^{\pi / 2}$$ f(x) = $$\int_{-\pi / 2}^{\pi / 2}$$ x3 = 0 ……………….(2)
[∵ $$\int_{-a}^a$$ f(x) = 0 if f(- x) = – f(x)]
Let g(x) = x cos x
∴ g(- x) = – x cos (- x)
= – x cos x
= – g(x)
∴ g(x) be an odd function.
$$\int_{-\pi / 2}^{\pi / 2}$$ g(x) dx = 0
⇒ $$\int_{-\pi / 2}^{\pi / 2}$$ x cos x dx = 0 ……………(3)
Let h(x) = tan5 x
⇒ h (- x) = [tan (- x)]5
= – tan5 x
= – h(x)
∴ $$\int_{-\pi / 2}^{\pi / 2}$$ h(x) dx = 0
⇒ $$\int_{-\pi / 2}^{\pi / 2}$$ tan5 x dx = 0 ……………….(4)
Using eqn. (2), (3) and (4) in eqn. (1) ; we get
I = 0 + 0 + 0 + x$$]_{-\pi / 2}^{\pi / 2}$$
= $$\frac{\pi}{2}+\frac{\pi}{2}$$
= π

(ii) Let I = $$\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}$$ dx …………….(1)

Question 45 (old).
(i) $$\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)$$ dx
Solution:
Let I = $$\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)$$ dx
= $$\int_0^1 \tan ^{-1}\left(\frac{x+(x-1)}{1-x(x-1)}\right)$$ dx
= $$\int_0^1$$ [tan-1 x + tan-1 (x – 1)] dx
I = $$\int_0^1$$ [tan-1 x + tan-1 (x – 1)] dx …………….(1)
∴ I = $$\int_0^1$$ [tan-1 (1 – x) + tan-1 (1 – x – 1)] dx
[∵ $$\int_0^a$$ f(x) dx = $$\int_0^a$$ f (a – x) dx]
⇒ I = $$\int_0^1$$ [tan-1 (1 – x) + tan-1 (- x)] dx
⇒ I = $$\int_0^1$$ [- tan-1 (x – 1) – tan-1 x] dx …………….(2)
[∵ tan-1 (- x) = – tan-1 x]
On adding (1) and (2) ; we have
2I = 0
⇒ I = 0.