OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(g)

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S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(g)

Question 1.
(i) \(\int \frac{d x}{4+5 \cos x}\)
(ii) \(\int \frac{d x}{12+12 \cos x}\)
(iii) \(\int \frac{d \theta}{4 \cos \theta-1}\)
(iv) \(\int \frac{d \theta}{1+2 \cos \theta}\)
Solution:
(i) \(\int \frac{d x}{4+5 \cos x}\)
Put tan \(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\)\(\frac { 1 }{ 2 }\)dx = dt
⇒ \(d x \frac{2 d t}{1+t^2}\)

(ii) \(\int \frac{d x}{12+12 \cos x}\)
Let I = \(\int \frac{d x}{12+12 \cos x}\) = \(\int \frac{d x}{12 \times 2 \cos ^2 \frac{x}{2}}\)
= \(\frac { 1 }{ 24 }\)\(\int \sec ^2 \frac{x}{2} d x\) = \(\frac { 1 }{ 24 }\)\(\frac{\tan \frac{x}{2}}{\frac{1}{2}}\) + C
= \(\frac { 1 }{ 12 }\)tan\(\frac { x }{ 2 }\) + C

(iii) \(\int \frac{d \theta}{4 \cos \theta-1}\)
Let I = \(\int \frac{d \theta}{4 \cos \theta-1}\)
Put tan \(\frac { θ }{ 2 }\) = t ⇒ sec²\(\frac { θ }{ 2 }\)\(\frac { 1 }{ 2 }\)dθ = dt
⇒ dθ = \(\frac{2 d t}{1+t^2}\)
∴cos θ = \(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\) = \(\frac{1-t^2}{1+t^2}\)

(iv) \(\int \frac{d \theta}{1+2 \cos \theta}\)
Put tan \(\frac { θ }{ 2 }\) = t ⇒ sec²\(\frac { θ }{ 2 }\) . \(\frac { 1 }{ 2 }\) dθ = dt
⇒ dθ = \(\frac{2 d t}{1+t^2}\)
& cos θ = \(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\) = \(\frac{1-t^2}{1+t^2}\)

Question 2.
(i) \(\int \frac{d x}{4-3 \sin x}\)
(ii) \(\int \frac{d x}{4+5 \sin x}\)
(iii) \(\int \frac{d \theta}{1-2 \sin \theta}\)
(iv) \(\int \frac{d x}{1-3 \sin x}\)
Solution:
(i) Let I = \(\int \frac{d x}{4-3 \sin x}\)
Put tan \(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\) . \(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
& tan x = \(\frac{2 \tan x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 d t}{1+t^2}\)

(ii) Let I = \(\int \frac{d x}{4+5 \sin x}\)
Put tan\(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\) . \(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 t}{1+t^2}\)
& sin x = \(\frac{2 \tan x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 d t}{1+t^2}\)

(iii) Let I = \(\int \frac{d \theta}{1-2 \sin \theta}\)
Put tan \(\frac { θ }{ 2 }\) = t ⇒ sec²\(\frac { θ }{ 2 }\) . \(\frac { 1 }{ 2 }\) dθ = dt
⇒ dθ = \(\frac{2 d t}{1+t^2}\)
& sin θ = \(\frac{2 \tan \theta / 2}{1+\tan ^2 \theta / 2}\) = \(\frac{2 t}{1+t^2}\)

(iv) Let I = \(\int \frac{d x}{1-3 \sin x}\)
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 = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 d t}{1+t^2}\)
∴ I = \(\int \frac{\frac{2}{1+t^2} d t}{1-3\left(\frac{2 t}{1+t^2}\right)}\) = \(\int \frac{2 d t}{t^2+1-6 t}\)
= \(2 \int \frac{d t}{t^2-6 t+9-8}\)
= \(2 \int \frac{d t}{(t-3)^2-(\sqrt{8})^2}\)
= 2 × \(\frac{1}{2 \sqrt{8}}\)log\(\left|\frac{t-3-\sqrt{8}}{t-3+\sqrt{8}}\right|\) + C
= \(\frac{1}{2 \sqrt{2}}\) log \(\left|\frac{\tan \frac{x}{2}-3-2 \sqrt{2}}{\tan \frac{x}{2}-3+2 \sqrt{2}}\right|\) + C

Question 3.
(i) \(\int \frac{d x}{2 \sin x-3 \cos x}\)
(ii) \(\int \frac{d x}{3 \cos x+4 \sin x}\)
(iii) \(\int \frac{d x}{\sin x-\cos x}\)
Solution:
(i) Let I = \(\int \frac{d x}{2 \sin x-3 \cos x}\)
Put tan \(\frac{x}{2}\) = t ⇒ sec²\(\frac { x }{ 2 }\) . \(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
sin x = \(\frac{2 \tan x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
& cos x = \(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\) = \(\frac{1-t^2}{1+t^2}\)
∴ I = \(\int \frac{\frac{2 d t}{1+t^2}}{2\left(\frac{2 t}{1+t^2}\right)-3\left(\frac{1-t^2}{1+t^3}\right)}\)
= \(\int \frac{2 d t}{4 t-3+3 t^2}\) = \(\frac { 2 }{ 3 }\)\(\)
= \(\frac { 2 }{ 3 }\)\(\int \frac{d t}{t^2+\frac{4}{3} t+\frac{4}{9}-\frac{4}{9}-1}\)
= \(\frac { 2 }{ 3 }\)\(\int \frac{d t}{\left(t-\frac{2}{3}\right)^2-\left(\frac{\sqrt{13}}{3}\right)^2}\)

(ii) Let I = \(\int \frac{d x}{3 \cos x+4 \sin x}\)
Put tan\(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\) . \(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)

(iii) Let I = \(\int \frac{d x}{\sin x-\cos x}\)
Put tan\(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\) . \(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
∴ sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
& cos x = \(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\) = \(\frac{1-t^2}{1+t^2}\)
∴ I = \(\int \frac{\frac{2 d t}{1+t^2}}{\frac{2 t}{1+t^2}-\frac{1-t^2}{1+t^2}}\)
= \(\int \frac{2 d t}{2 t-1+t^2}\) = \(\int \frac{2 d t}{t^2+2 t+1-2}\)
= \(\int \frac{2 d t}{(t+1)^2-(\sqrt{2})^2}\)
= \(2 \int \frac{d t}{u^2-(\sqrt{2})^2}\)
[Put t + 1 = u ⇒ dt = du]
= 2 × \(\frac{1}{2 \sqrt{2}}\) log \(\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|\) + C
= \(\frac{1}{\sqrt{2}}\) log\(\left|\frac{\tan \frac{x}{2}+1-\sqrt{2}}{\tan \frac{x}{2}+1+\sqrt{2}}\right|\) + C

Question 4.
(i) \(\int \frac{d x}{3+2 \sin x+\cos x}\)
(ii) \(\int \frac{d x}{1-\sin x+\cos x}\)
(iii) \(\int \frac{d x}{1+\sin x+\cos x}\)
Solution:
(i) Let I = \(\int \frac{d x}{3+2 \sin x+\cos x}\)
Put tan\(\frac{x}{2}\) = t ⇒ sec²\(\frac{x}{2}\) \(\frac{1}{2}\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
& cos x = \(\frac{1-\tan ^2 x / 2}{1-\tan ^2 x / 2}\) = \(\frac{1-t^2}{1+t^2}\)
∴ I = \(\int \frac{\frac{2 d t}{1+t^2}}{3+2\left(\frac{2 t}{1+t^2}\right)+\frac{1-t^2}{1+t^2}}\)
= \(\int \frac{2 d t}{3\left(1+t^2\right)+4 t+1-t^2}\) = \(\int \frac{2 d t}{2 t^2+4 t+4}\)
= \(\int \frac{2 d t}{2\left(t^2+2 t+2\right)}\) = \(\int \frac{2 d t}{(t+1)^2+1}\)
= tan^-1 (t + 1) + C
= tan^-1 (1 + tan\(\frac{x}{2}\)) + C

(ii) Let I = \(\int \frac{d x}{1-\sin x+\cos x}\)
Put tan\(\frac{x}{2}\) = t ⇒ sec²\(\frac{x}{2}\) \(\frac{1}{2}\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
& cos x = \(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\) = \(\frac{1-t^2}{1+t^2}\)
∴ I = \(\int \frac{\frac{2 d t}{1+t^2}}{1-\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}}\)
= \(\int \frac{2 d t}{1+t^2-2 t+1-t^2}\)
= \(\int \frac{d t}{1-t}\) = \(\frac{\log |1-t|}{-1}\) + C
= – log | 1 – tan\(\frac{x}{2}\) | + C

(iii) Let I = \(\int \frac{d x}{1+\sin x+\cos x}\)
Put tan\(\frac{x}{2}\) = t ⇒ sec²\(\frac{x}{2}\) \(\frac{1}{2}\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
sin x = \(\frac{2 \tan x / 2}{1+\tan ^2 \frac{x}{2}}\) = \(\frac{2 t}{1+t^2}\)
& cos x = \(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\) = \(\frac{1-t^2}{1+t^2}\)
∴ I = \(\int \frac{\frac{2 d t}{1+t^2}}{1+\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}}\)
= \(\int \frac{2 d t}{1+t^2+2 t+1-t^2}\) = \(\int \frac{d t}{1+t}\)
= log |t + 1| + C
= log | 1 + tan\(\frac{x}{2}\) | + C

Question 5.
(i) \(\int \frac{2 \sin \theta+\cos \theta}{7 \sin \theta-5 \cos \theta} d \theta\)
(ii) \(\int \frac{\cos x+3 \sin x-5}{3(1-\sin x)-\cos x} d x\)
Solution:
(i) Let I = \(\int \frac{2 \sin \theta+\cos \theta}{7 \sin \theta-5 \cos \theta} d \theta\)
Let numerator = l(deno) + m\(\frac{d}{d \theta}\)(deno)
⇒ 2 sinθ + cosθ = l(7sinθ – 5cosθ) + m\(\frac{d}{d \theta}\)(7 sinθ – 5cosθ)
⇒ 2 sinθ + cosθ = l(7sinθ – 5cosθ) + m(7cosθ + 5 sinθ) …(i)
Comparing the coeff. of sinθ & cosθ on both sides; we have
2 = 7l + 5m ⇒ 7l + 5m – 2 = 0 & 1 = – 5l + 7m ⇒ – 5l + 7m – 1 = 0
∴ \(\frac{l}{-5+14}\) = \(\frac{m}{10+7}\) = \(\frac{9}{49+25}\) ⇒ l = \(\frac{9}{74}\) & m = \(\frac{17}{74}\)
∴ from (i); we have
2 sinθ + cosθ = \(\frac{9}{74}\)(7sinθ – 5cosθ) + \(\frac{17}{74}\)(7 cosθ + 5 sinθ)

(ii) Let I = \(\int \frac{\cos x+3 \sin x-5}{3(1-\sin x)-\cos x} d x\)
Numerator = l(deno) + m\(\frac{d}{dx}\) l(deno) + n
i.e. cos x + 3 sin x – 5 = l (3 – 3 sin x – cos x) l + m(- 3 cos x + sin x) + n …(1)
Coeff. of sin x : 3 = – 3l + m …(2)
Coeff. of cos x : 1 = – 1 – 3 m …(3)
Constant term; – 5 = 3 l + n …(4)
Multiply eqn. 2 by 3 + eqn. (3); we have
1 + 9 = – 9l – l ⇒ l = –\(\frac{10}{10}\) = -1
∴ from (2); m = 3 + 3l = 3 – 3 = 0
∴ from (4); n = – 5 – 3l = – 5 + 3 = – 2
Thus from eqn. (1); we have
cos x + 3 sin x – 5 = – 1(3 – 3 sin x – cos x) + 0 (-3 cos x + sin x) – 2

Question 6.
(i) \(\int \frac{d x}{4 \sin ^2 x+5 \cos ^2 x}\)
(ii) \(\int \frac{d x}{2 \cos ^2 x+\sin ^2 x}\)
(iii) \(\int \frac{d x}{1+3 \sin ^2 x}\)
(iv) \(\int \frac{d x}{\cos 2 x+3 \sin ^2 x}\)
(v) \(\int \frac{d x}{9 \cos ^2 x+8 \sin ^2 x+3}\)
(vi) \(\int \frac{d x}{(2 \sin x+3 \cos x)^2}\)
Solution:
(i) Let I = \(\int \frac{d x}{4 \sin ^2 x+5 \cos ^2 x}\)
Divide Numerator and denomenator by cos² x; we have

(ii) Let I = \(\int \frac{d x}{2 \cos ^2 x+\sin ^2 x}\)
Divide numerator and deno by cos²x; we have
= \(\int \frac{\sec ^2 x d x}{2+\tan ^2 x}\)
Put tan x = t ⇒ sec²x dx = dt
= \(\int \frac{d t}{2+t^2}\) = \(\int \frac{d t}{t^2+(\sqrt{2})^2}\)
= \(\frac{1}{\sqrt{2}}\) tan^-1\(\left(\frac{t}{\sqrt{2}}\right)\) + C
= \(\frac{1}{\sqrt{2}}\)tan^-1\(\left(\frac{\tan x}{\sqrt{2}}\right)\) + C

(iii) Let I = \(\int \frac{d x}{1+3 \sin ^2 x}\)
Divide Numerator & deno by cos²x ; we have
= \(\int \frac{\sec ^2 x d x}{\sec ^2 x+3 \tan ^2 x}\) = \(\int \frac{\sec ^2 x d x}{1+4 \tan ^2 x}\)
Put tan x = t ⇒ sec²x dx = dt

(iv) Let I = \(\int \frac{d x}{\cos 2 x+3 \sin ^2 x}\)
= \(\int \frac{d x}{\cos ^2 x-\sin ^2 x+3 \sin ^2 x}\)
= \(\int \frac{d x}{\cos ^2 x+2 \sin ^2 x}\)
Divide Numerator and deno by cos²x; we have
= \(\int \frac{\sec ^2 x d x}{1+2 \tan ^2 x}\)
Put tan x = t ⇒ sec²x dx = dt

(v) Let I = \(\int \frac{d x}{9 \cos ^2 x+8 \sin ^2 x+3}\)
Divide numerator and deno. by cos²x; we have

(vi) Let I = \(\int \frac{d x}{(2 \sin x+3 \cos x)^2}\)
Divide Numerator & denomirator by cos²x; we get

Question 7.
(i) \(\int \frac{d x}{4+3 \sin 2 x}\)
(ii) \(\int \frac{d x}{\sin 2 x+4}\)
Solution:
(i) Let I = \(\int \frac{d x}{4+3 \sin 2 x}\)
= \(\int \frac{d x}{4+6 \sin x \cos x}\)
Divide Numerator & denomirator by cos²x; we get

(ii) Let I = \(\int \frac{d x}{\sin 2 x+4}\) = \(\int \frac{d x}{2 \sin x \cos x+4}\)
Divide numerator & denomirator by cos²x, we have

Question 8.
\(\int \frac{\sin 2 x}{\left(\sin ^4 x+\cos ^4 x\right)} d x\)
Solution:
Let I = \(\int \frac{\sin 2 x d x}{\sin ^4 x+\cos ^4 x}\);
Divide Numerator & denominator by cos⁴x; we have
= \(\int \frac{2 \tan x \sec ^2 x d x}{\tan ^4 x+1}\)
put tan²x = t ⇒ 2 tan x sec²x dx = dt
= \(\int \frac{d t}{t^2+1^2}\) = \(\frac{1}{1}\) tan^-1 t + c
= tan^-1 (tan²x) + C