ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.6

Parents can use ML Aggarwal Class 12 ISC Solutions Chapter 8 Integrals Ex 8.6 to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.6

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 8) integrals:

Question 1.
(i) ∫ tan (2 – 3x) dx
(ii) ∫ cot (7 – 4x) dx
Solution:
(i) Let I = ∫ \(\frac{\sin (2-3 x)}{\cos (2-3 x)}\) dx
put cos (2 – 3x) = t
⇒ – sin (2 – 3x) (- 3) dx = dt
= ∫ \(\frac{d t}{3}\)
= \(\frac{1}{3}\) log |t| + C
= \(\frac{1}{3}\) log |cos (2 – 3x)| + C

(ii) Let I = ∫ cot (7 – 4x) dx
= ∫ \(\frac{\cos (7 x-4)}{\sin (7 x-4)}\) dx
put sin (7x – 4) = t
⇒ cos (7x – 4) . 7 dx = dt
= ∫ \(\frac{d t}{7 t}\)
= \(\frac{1}{7}\) log |t| + C
= \(\frac{1}{7}\) log |sin (7x – 4)| + C

Question 2.
(i) ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
(ii)∫ \(\frac{1}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
= \(\frac{1}{2} \int \frac{\sec ^2 x d x}{\tan x}\)
= \(\frac{1}{2}\) log |tan x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

Aliter:
Let I = ∫ \(\frac{\sec ^2 x d x}{2 \tan x}\)
= ∫ \(\frac{d x}{2 \sin x \cos x}\)
= ∫ \(\frac{d x}{\sin 2 x}\)
= ∫ cosec 2x dx
= \(\frac{\log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C

(ii) Let I = ∫ \(\frac{d x}{\sin x \cos x}\)
= ∫ \(\frac{2 d x}{\sin 2 x}\)
= 2 ∫ cosec 2x dx
= \(\frac{2 \log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C
= log |cosec 2x – cot 2x| + C

Question 3.
(i) ∫ \(\frac{\cos 2 x}{\sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
(iii) ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos 2 x}{\sin x}\) dx
= ∫ \(\frac{\left(1-2 \sin ^2 x\right)}{\sin x}\) dx
= ∫ cosec x dx – 2 ∫ sin x dx + C
= log |cosec x – cot x| + 2 cos x + C

(ii) Let I = ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
= ∫ \(\frac{\sin 2 x d x}{2 \sin 2 x \cos 2 x}\) dx
= \(\frac{1}{2}\) ∫ sec 2x dx
= \(\frac{1}{2} \frac{\log |\sec 2 x+\tan 2 x|}{2}\) + C
= \(\frac{1}{4}\) log |sec 2x + tan 2x| + C

(iii) Let I = ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
= \(\int \frac{\sin x}{\cos x} d x-\int \frac{\cos x}{\sin x} d x\)
= – \(\int \frac{-\sin x}{\cos x} d x-\int \frac{\cos x d x}{\sin x}\)
= – log |cos x| – log |sin x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= log |sec x| – log |sin x| + C
[∵ a log b = log b^q]

Question 4.
(i) ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
(ii) ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
= ∫ \(\frac{\sin (x-\alpha+\alpha) d x}{\sin (x-\alpha)}\)
= ∫ cos α dx + sin α ∫ \(\frac{\cos (x-\alpha)}{\sin (x-\alpha)}\) dx
= x cos α + sin α log |sin (x – α)| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos (x+\alpha-\alpha)}{\cos (x+\alpha)}\) dx
= ∫ \(\left[\frac{\cos (x+\alpha) \cos \alpha-\sin (x+\alpha) \sin \alpha}{\cos (x+\alpha)}\right]\) dx
= ∫ cos α dx – sin α ∫ \(\frac{\sin (x+\alpha) d x}{\cos (x+\alpha)}\)
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= x cos α + sin α log |cos (x + α)| + C

Question 5.
(i) ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)
(ii) ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)

(ii) Let I = ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)

Question 6.
(i) ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)
(ii) ∫ tan³ x dx (ISC 2016)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)

(ii) Let I = ∫ tan³ x dx
= ∫ tan x (sec² – 1) dx
= ∫ tan x (sec² x dx) – ∫ tan x dx
= \(\frac{\tan ^2 x}{2}\) + log |cos x| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ 1]

Question 7.
(i) ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx
(ii) ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx

(ii) Let I = ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
= ∫ \(\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin x \cos ^2 x}\)
= ∫ tan x sec x dx + ∫ cosec x dx
= sec x + log |cosec x – cot x| + C

Question 8.
(i) ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
(ii) ∫ \(\frac{\sin x+\cos x}{1+\sin 2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
= ∫ \(\frac{d x}{\sqrt{1-\cos \left(\frac{\pi}{2}-x\right)}}\)
= ∫ \(\frac{d x}{\sqrt{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}\)
= \(\frac{1}{\sqrt{2}} \int \ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
= \(\frac{1}{\sqrt{2}} \frac{\log \left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|}{-\frac{1}{2}}\) + C
= – √2 log \(\left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|\) + C

Aliter:

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

Question 9.
∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx = ax + b log |sin (x – \(\frac{\pi}{4}\))| + C, find the values of a and b.
Solution:
Let I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx

Also given I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx
= ax + b log |sin (x – \(\frac{\pi}{4}\))| + C …………(2)
∴ From (1) and (2) ; we have
a = 1 ;
b = 1.