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

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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) = \(\frac{2}{3}\) x^3/2 + 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) = \(\frac{2}{3}\) x^3/2 + \(\frac{4}{3}\)

Question 1 (old).
(i) ∫ \(\frac{x}{\sqrt{1-x^2}}\) dx
(ii) ∫ x² e^x3 dx
(iii) ∫ e^ex e^x 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 e^x3 dx
put x³ = t
⇒ 3x² dx = dt
= ∫ e^t \(\frac{d t}{3}\)
= \(\frac{e^{x^3}}{3}\) + C

(iii) put e^x = t
e^x dx = dt
∴ I = ∫ e^ex e^x dx
= ∫ e^t dt
= e^t + C
= e^ex + 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 + 10^x = t
⇒ (10x⁹ + 10^x 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 + 10^x| + 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) ∫ e^{cot x} cosec² x dx
Solution:
(i) put e^x = t
⇒ e^x 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 |e^x + \(\sqrt{e^{2 x}-4}\)| + C

(ii) Let I = ∫ e^{cot x} cosec² x dx
put cos x = t
⇒ – cosec² x dx = dt
∴ I = ∫ e^t (- dt)
= – e^t + C
= – e^{cot 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 = ∫ e^t dt
= e^t + 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 = e^t
⇒ dx = e^t dt
∴ I = ∫ e^t/2 t² e^t 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 e^x = t
⇒ e^x 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 x^1/4 = t
⇒ x = t⁴
⇒ dx = 4t³ dt

= 2t² + 4t + 4 log |t – 1| + C
= 2√x + 4x^1/4 + 4 log |x^1/4 – 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) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x 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 = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x 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) \(\int_0^1\) e^{2 – 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\(\int_0^1\) e^{2 – 3x} dx with \(\int_a^b\) f(x) dx
Here, f(x) = e^2-3x;
a = 0;
b = 1;
nh = b – a = 1 – 0 = 1
∴ f(a) = f(0) = e²
f(a + h) = f(h) = e^{2 – 3h}
f(a + 2h) = f(2h) = e^{2 – 6h}
……………………………
……………………………
\(f(a+\overline{n-1} h)\) = f((n – 1)h)
= e^{2 – 3 (n – 1) h}
∴ \(\int_0^1\) e^{2 – 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 + e^{2 – 3h} + e^{2 – 6h} + ……. + e^{2 – 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.