OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Ex 16(b)

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S Chand Class 12 ICSE Maths Solutions Chapter 16 Definite Integrals Ex 16(b)

Question 1.
(i) \(\int_1^2 \frac{3 x}{9 x^2-1} d x\)
(ii) \(\int_0^1 \frac{x^5}{1+x^6} d x\)
(iii) \(\int_0^1 \frac{x}{1+x^4} d x\)
(iv) \(\int_0^1 \frac{e^{2 x}}{1+x^{4 x}} d x\)
(v) \(\int_1^2 3 x \sqrt{5-x^2} d x\)
(vi) \(\int_1^2 \frac{e^{-\frac{1}{x}}}{x^2} d x\)
(vii) \(\int_0^a x \sqrt{a^2-x^2} d x\)
(viii) \(\int_0^{(\pi / 2)^{1 / 3}} x^2 \sin x^3 d x\)
Solution:
(i) Let I = \(\int_1^2 \frac{3 x d x}{9 x^2-1}\)
Put 9x² – 1 = t ⇒ 18xdx = dt
⇒ xdx = \(\frac { 1 }{ 18 }\)dt
when x = 1 ⇒ t = 8;
when x = 2 ⇒ t = 35
= \(\left.\int_8^{35} \frac{d t}{6 . t}=\frac{1}{6} \log |t|\right]_8^{35}\)
= \(\frac { 1 }{ 6 }\)[log 35 – log 8]
= \(\frac { 1 }{ 6 }\) log \(\frac { 35 }{ 8 }\)

(ii) Let I = \(\int_0^1 \frac{x^5 d x}{1+x^6}\);
Put 1 + x² = t ⇒ 6x⁵dx = dt
when x = 0 ⇒ t = 1;
when x= 1 ⇒ t = 2
= \(\left.\int_1^2 \frac{d t}{6 . t}=\frac{1}{6} \log |t|\right]_1^2\)
= \(\frac { 1 }{ 6 }\) [log 2 – log 1]
= \(\frac { 1 }{ 6 }\) log 2

(iii) \(\int_0^1 \frac{x d x}{1+x^4}\)
Put x² = t ⇒ 2xdx = dt
when x = 0 ⇒ t = 0;
when x = 1 ⇒ t =1
= \(\left.\int_0^1 \frac{d t}{2\left(1+t^2\right)}=\frac{1}{2} \tan ^{-1} t\right]_0^1\)
= \(\frac { 1 }{ 2 }\)[tan^-1 1 – tan^-1 0]
= \(\frac { 1 }{ 2 }\) \(\left[\frac{\pi}{4}-0\right]\) = \(\frac{\pi}{8}\)

(iv) Let I = \(\int_0^1 \frac{e^{2 x}}{1+e^{4 x}} d x\)
Put e^2x = t ⇒ 2e^2xdx = dt
when x = 0 ⇒ t = e⁰ = 1;
when x = 1 ⇒ t = e²

(v) Let I = \(\int_1^2 3 x \sqrt{5-x^2} d x\)
Put x² = t ⇒ 2xdx = dt
when x = 1 ⇒ t = 1;
when x =2 ⇒ t = 4

(vi) Let I = \(\int_1^2 \frac{e^{-\frac{1}{x}}}{x^2} d x\)
Put \(\frac { 1 }{ x }\) = t ⇒ –\(\frac{1}{x^2}\)dx = dt
when x = 1 ⇒ t = 1;
when x = 2 ⇒ t = \(\frac { 1 }{ 2 }\)
∴ I = \(\left.\int_1^{1 / 2} e^{-t}(-d t)=e^{-t}\right]_1^{1 / 2}\)
= (e^-1/2 – e^-1)
= –\(\frac { 1 }{ e }\) + \(\frac{1}{\sqrt{e}}\) = \(\frac{\sqrt{e}-1}{e}\)

(vii) \(\int_0^a x \sqrt{a^2-x^2} d x\)
Put a² – x² = t ⇒ -2xdx = dt
⇒ xdx = –\(\frac { dt }{ 2 }\)
when x = 0 ⇒ t = a²;
when x = a ⇒ t = a² – a² = 0
∴ I = \(\left.\int_{a^2}^0 \sqrt{t} \frac{d t}{-2}=-\frac{1}{2} \frac{t^{3 / 2}}{3 / 2}\right]_{a^2}^0\)
= –\(\frac { 1 }{ 3 }\)[0 – (a²)^3/2] = \(\frac{a^3}{3}\)

(viii) \(\int_0^{(\pi / 2)^{1 / 3}} x^2 \sin x^3 d x\)
Put x³ = t ⇒ 3x²dx =dt
when x = 0 ⇒ t = 0;
when x = \(\left(\frac{\pi}{2}\right)^{1 / 3}\) ⇒ t = \(\frac{\pi}{2}\)

Question 2.
(i) \(\int_0^{\pi / 2} \frac{\cos x}{1+\sin ^2 x} d x\)
(ii) \(\int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} d x\)
(iii) \(\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x} d x\)
(iv) \(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x} d x\)
(v) \(\int_0^{\pi / 2} \frac{d x}{4 \sin ^2 x+5 \cos ^2 x}\)
(vi) \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x\)
(vii) \(\int_0^{\pi / 4} \frac{\tan ^3 x}{1+\cos 2 x} d x\)
(viii) \(\int_1^e \frac{\cos (\log x)}{x} d x\)
(ix) \(\int_0^{\pi / 2} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta\)
(x) \(\int_0^{\pi / 4} 2 \tan ^3 x d x\)
(xi) \(\int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x\)
(xii) \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x\)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\cos x d x}{1+\sin ^2 x}\)
Put sin x = t ⇒ cos xdx = dt
when x = 0 ⇒ t = 0;
when x = \(\frac{\pi}{2}\) ⇒ t = 1
∴ I = \(\left.\int_0^1 \frac{d t}{1+t^2}=\tan ^{-1} t\right]_0^1\)
= tan^-1 1 – tan^-1 0 = \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin x d x}{1+\cos ^2 x}\)
Put cos x = t ⇒ -sin xdx = dt
when x = 0 ⇒ t = 1;
when x = \(\frac{\pi}{2}\) ⇒ t = 0
∴ I = \(\left.-\int_1^0 \frac{d t}{1+t^2}=-\tan ^{-1} t\right]_1^0\)
= -[tan^-1 0 – tan^-1 1]
= -[0 – \(\frac{\pi}{4}\)] = \(\frac{\pi}{4}\)

(iii) \(\int_0^{\pi / 3} \frac{\sec x \tan x d x}{1+\sec ^2 x}\)
Put sec = t ⇒ secx tanx dx = dt
when x = 0 ⇒ t = 1;
when x = \(\frac{\pi}{3}\) ⇒ t = sec\(\frac{\pi}{3}\) = 2
∴ I = \(\left.\int_1^2 \frac{d t}{1^2+t^2}=\tan ^{-1} t\right]_1^2\)
= tan^-1 2 – tan^-1 1
= tan^-1 2 – \(\frac{\pi}{4}\)

(iv) Let I = \(\int_0^{\pi / 2} \frac{\sin x \cos x d x}{1+\sin ^4 x}\)
Put sin²x = t ⇒ \(\frac{1}{1+x^2}\) dx = dt
when x = 0 ⇒ t = tan^-10 = 0;

(v) Let I = \(\int_0^{\pi / 2} \frac{d x}{4 \sin ^2 x+5 \cos ^2 x}\)
Divide Numerator and deno by cos²x, we have
= \(\int_0^{\pi / 2} \frac{\sec ^2 x d x}{4 \tan ^2 x+5}\)
put tan x = t ⇒ sec²x dx = dt
when x = 0 ⇒ t = 0;

(vi) Let I = \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x\)
Put tan^-1x = t ⇒ \(\frac{1}{1+x^2}\) dx = dt
when x = 0 ⇒ t = tan^-10 = 0;
when x = 1 ⇒ t = tan^-11 = \(\frac{\pi}{4}\)

(vii) Let I = \(\int_0^{\pi / 4} \frac{\tan ^3 x}{1+\cos 2 x} d x\)
= \(\int_0^{\pi / 4} \frac{\tan ^3 x}{2 \cos ^2 x} d x\)
= \(\frac{1}{2}\)\(\int_0^{\pi / 4} \tan ^3 x \sec ^2 x d x\)
Put tanx = t ⇒ sec²x dx = dt
when x = 0 ⇒ t = 0;
when x = \(\frac{\pi}{4}\) ⇒ t = tan\(\frac{\pi}{4}\) = 1
∴ I = \(\left.\frac{1}{2} \int_0^1 t^3 d t=\frac{1}{2} \frac{t^4}{4}\right]_0^1\)
= \(\frac{1}{8}\)(1 – 0) = \(\frac{1}{8}\)

(viii) Let I = \(\int_1^e \frac{\cos (\log x)}{x} d x\)
Put log x = t ⇒ \(\frac{1}{4}\) dx = dt
when x = e ⇒ t = log e = 1;
when x = 1 ⇒ t = log 1 = 0
∴ I = \(\left.\int_0^1 \cos t d t=\sin t\right]_0^1\)
= sin 1 – sin 0 = sin 1

(ix) Let I = \(\int_0^{\pi / 2} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta\)
= \(\int_0^{\pi / 2} \sqrt{\cos \theta}\left(1-\cos ^2 \theta\right) \sin \theta d \theta\)
Put cosθ = t ⇒ -sinθdθ = dt
when θ = 0 ⇒ t = 1;
when θ = \(\frac{\pi}{2}\) ⇒ t = 0
∴ I = \(\int_1^0 \sqrt{t}\left(1-t^2\right)(-d t)\)
= \(-\left[\frac{2 t}{3}^{3 / 2}-\frac{2}{7} t^{7 / 2}\right]_1^0\)
= \(-\left[(0-0)-\left(\frac{2}{3}-\frac{2}{7}\right)\right]\)
= \(\frac{14-6}{21}\) = \(\frac{8}{21}\)

(xi) Let I = \(\int_0^{\pi / 2} \frac{\sin x \cos x d x}{\cos ^2 x+3 \cos x+2}\)
Put cosx = t ⇒ -sinx dx = dt
when x = 0 ⇒ t = 1;
when x = \(\frac{\pi}{2}\) ⇒ t = 0

(xii) Let I = \(\int_0^{\pi / 2} \frac{\cos x d x}{(1+\sin x)(2+\sin x)}\)
Put sinx = t ⇒ cosx dx = dt; when x = 0 ⇒ t = 0; when x = \(\frac{\pi}{2}\) ⇒ t = 1

Question 3.
(i) \(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x\)
(ii) \(\int_0^1 x \sqrt{\frac{1-x}{1+x}} d x\)
(iii) \(\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x\)
(iv) \(\int_0^2 \frac{6 x+3}{x^2+4} d x\)
Solution:
(i) Let I = \(\int_0^1 \sqrt{\frac{1-x}{1+x}} d x\)
Put x = cosθ ⇒ dx = -sinθ dθ
when x = 0 ⇒ θ = 1;
when x = 1 ⇒ θ = 0

(ii) Let I = \(\int_0^1 x \sqrt{\frac{1-x^2}{1+x^2}} d x\)
Put x² = cosθ ⇒ 2xdx = -sinθdθ
when x = 0 ⇒ θ = π/2;
when x = 1 ⇒ θ = 0

(iv) Let I = \(\int_0^2 \frac{6 x+3}{x^2+4} d x\)
= \(\int_0^2 \frac{6 x d x}{x^2+4}\) + \(3 \int_0^2 \frac{d x}{x^2+2^2}\)
Put x² + 4 = t ⇒ 2xdx = dt
when x = 0 ⇒ t = 4;
when x= 2 ⇒ t = 8

Question 4.
(i) \(\int_0^{\pi / 2} \frac{d x}{3+2 \cos x}\)
(ii) \(\int_0^{\pi / 2} \frac{d x}{7+4 \sin x}\)
(iii) \(\int_0^{\pi / 2} \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x\)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{d x}{3+2 \cos x}\)
Put tan \(\frac { x }{ 2 }\) = t ⇒ sec² \(\frac { x }{ 2 }\)\(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
∴ cos x = \(\frac{1-\tan ^2 x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{1-t^2}{1+t^2}\)
when x = 0 ⇒ t = 0;
when x = \(\frac{\pi}{2}\) ⇒ t = tan\(\frac{\pi}{4}\) = 1

(ii) Let I = \(\int_0^{\pi / 2} \frac{d x}{7+4 \sin x}\)
Put tan\(\frac{x}{2}\) = t ⇒ sec² \(\frac { x }{ 2 }\)\(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
sin x = \(\frac{2 \tan ^2 x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
when x = 0 ⇒ t = 0;
when x = \(\frac{\pi}{2}\) ⇒ t = tan\(\frac{\pi}{4}\) = 1

(iii) Let I = \(\int_0^{\pi / 2} \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x\)
Divide Numerator and denominator by cos²; we have
= \(\int_0^{\pi / 2} \frac{2 \tan x \sec ^2 x d x}{\tan ^4 x+1} d x\)
Put tan²x = t ⇒ 2tanx sec²x dx = dt
when x = 0 ⇒ t = 0;
when x = \(\frac{\pi}{2}\) ⇒ t = → ∞
= \(\left.\int_0^{\infty} \frac{d t}{t^2+1^2}=\tan ^{-1} t\right]_0^{\infty}\)
= tan^-1 ∞ – tan^-1 0
= \(\frac{\pi}{2}\) – 0 = \(\frac{\pi}{2}\)

Question 5.
(i) \(\int_0^{\sqrt{2}} \sqrt{2-x^2} d x\)
(ii) \(\int_0^3 \sqrt{9-x^2} d x\)
(iii) \(\int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x\)
(iv) \(\int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}}\)
(v) \(\int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x\)
Solution:
(i) \(\int_0^{\sqrt{2}} \sqrt{2-x^2} d x\)
put x = √2 sinθ ⇒ dx = √2 cosθ dθ
when x = 0 ⇒ θ = 0;
when x = √2 ⇒ θ = \(\frac{\pi}{2}\)

(ii) \(\int_0^3 \sqrt{9-x^2} d x\) = \(\int_0^3 \sqrt{3^2-x^2} d x\)
= \(\left.\frac{x \sqrt{9-x^2}}{2}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_0^3\)
= \(\frac{3 \times 0}{2}\) + \(\frac{9}{2}\) sin^-1 1 – 0 = 0
= \(\frac{9}{2}\) × \(\frac{\pi}{2}\) = \(\frac{9 \pi}{4}\)

(iii) Let I = \(\int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x\)
put x = a sin θ ⇒ dx = a cosθ dθ
when x = 0 ⇒ θ = 0;
when x =a ⇒ sin θ = 1 ⇒ θ = \(\frac{\pi}{2}\)

(iv) Let I = \(\int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}}\)
put \(\sqrt{x+1}\) = t ⇒ x + 1 = t²
⇒ dx = 2tdt
when x = 8 ⇒ t =\(\sqrt{8+1}\) = 3
when x = 15 ⇒ t = \(\sqrt{15+1}\) = 4

(v) Let I = \(\int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x\)

Question 6.
(i) \(\int_0^1 \cos ^{-1} x d x\)
(ii) \(\int_0^1 \tan ^{-1} x d x\)
(iii) \(\int_0^1 x \sin ^{-1} x d x\)
(iv) \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Solution:
(i) Let I =\(\int_0^1 \cos ^{-1} x \cdot 1^{(1)} d x\)
= \(\left.\left(\cos ^{-1} x\right) x\right]_0^1-\int_0^1-\frac{1}{\sqrt{1-x^2}} x d x\)
= \(\left(\cos ^{-1} 1-0\right)+\int_0^1\left(1-x^2\right)^{\frac{-1}{2}} x d x\)
= \((0-0)-\frac{1}{2} \int_0^1\left(1-x^2\right)^{\frac{-1}{2}}(-2 x d x)\)
= \(\left.\left.-\frac{1}{2} \frac{\left(1-x^2\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_0^1=-\sqrt{1-x^2}\right]_0^1\)
= -[0 – 1] = 1

(ii) Let I = \(=\int_0^1 \tan ^{-1} x d x\)

(iii) Let I = \(\int_0^1 x \sin ^{-1} x d x\)

(iv) Let I = \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Put x = tanθ ⇒ dx = sec²θdθ
∴ θ = tan^-1x
when x = 0 ⇒ θ = 0; when x = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 7.
Evaluate : \(\int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x\)
Solution:

Question 8.
Prove that:
(i) \(\int_0^a \sin ^{-1} \sqrt{\left(\frac{x}{a+x}\right)} d x=a\left(\frac{\pi}{2}-1\right)\)
(ii) \(\int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\sqrt{2} \cdot \frac{\pi}{2}\)
Solution:
(i) Let I = \(\int_0^a \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\)
Put x = a tan²θ ⇒ dx = 2a tan θ sec²θ dθ
when x = 0 ⇒ θ = 0; when x= a ⇒ tan²θ = 1 ⇒ θ = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x\)
= \(\int_0^{\pi / 4}\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) d x\) = \(\int_0^{\pi / 4}\left(\frac{\tan x+1}{\sqrt{\tan x}}\right) d x\)
Put \(\sqrt{\tan x}\) = t ⇒ tanx = t² ⇒ sec²xdx = 2tdt
⇒ dx = \(\frac{2 t d t}{1+\tan ^2 x}\) = \(\frac{2 t d t}{1+t^4}\)
when x= 0 ⇒ t = 0; when x = \(\frac{\pi}{4}\) ⇒ t = 1
∴ I = \(\int_0^1\left(\frac{t^2+1}{t}\right)\left(\frac{2 t d t}{1+t^4}\right)\)
= \(2 \int_0^1\left(\frac{t^2+1}{1+t^4}\right) d t\)
Divive Numerator and deno minator by t²; we have
= \(2 \int_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\frac{1}{t^2}+t^2}\)
Put t – \(\frac{1}{4}\) = u ⇒ \(\left(1+\frac{1}{t^2}\right)\) dt = du
on squaring both sides, we have
t² + \(\frac{1}{t^2}\) – 2 = u²
when t = 0 ⇒ u → -∞
when t =1 ⇒ u = 0