ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.4

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.4

Very Short answer type questions (1 to 9):

Evaluate the following (1 to 21) integrals:

Question 1.
(i) ∫ x³ (1 + x⁴)³ dx
(ii) ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
Solution:
(i) Let I = ∫ x³ (1 + x⁴)³ dx
= \(\frac{1}{4}\) ∫ (1 + x⁴)x³ dx
= \(\frac{1}{4} \frac{\left(1+x^4\right)^4}{4}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(1+x^4\right)^4}{16}\) + C

(ii) Let I = ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
= ∫ (x² + 4x + 3)^1/2 (2x + 4) dx
= \(\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (x² + 4x + 3)^3/2 + C

Question 2.
(i) ∫ x (x² + 3)^3/2 dx
(ii) ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
Solution:
(i) ∫ x (x² + 3)^3/2 dx
= \(\frac{1}{2}\) ∫ (x² + 3)^3/2 dx
= \(\frac{1}{2} \frac{\left(x^2+3\right)^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{1}{5}\) (x² + 3)^5/2 + C

(ii) Let I = ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
= ∫ (2x² + x – 7)^{\(-\frac{1}{2}\)} (4x + 1) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(2 x^2+x-7\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2 \(\sqrt{2 x^2+x-7}\) + C

Question 3.
(i) ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
(ii) ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
Solution:
(i) Let I = ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
= ∫ (3x² + 2x – 5)^{– 3} (3x + 1) dx
= \(\frac{1}{2}\) ∫ (3x² + 2x – 5)^{– 3} (6x + 2) dx
= \(\frac{1}{2} \frac{\left(3 x^2+2 x-5\right)^{-3+1}}{(-3+1)}\) + C
= – \(\frac{1}{4}\) ∫ (3x² + 2x – 5)^{– 2} + C
= – \(\frac{1}{4\left(3 x^2+2 x-5\right)^2}\) + C

(ii) Let I = ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
put x² + 1 = t
⇒ 2x + dx = dt
= ∫ \(\frac{d t}{t^{1 / 3}}\)
= ∫ t^{\(-\frac{1}{3}\)} dt
= \(\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\) + C
= \(\frac{3}{2}\) t^2/3 + C
= \(\frac{3}{2}\) (x² + 1)^2/3 + C

Question 3 (old).
(ii) ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
Solution:
Let I = ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
= ∫ (x² + x – 3)^{\(\frac{1}{2}\)} 2 (2x + 1) dx
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{4}{3}\) (x² + x – 3)^3/2 + C

Question 4.
(i) ∫ e^x (a + be^x)⁶ dx
(ii) ∫ \(\frac{(\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ e^x (a + be^x)⁶ dx
= \(\frac{1}{b}\) ∫ (a + be^x)⁶ be^x dx
= \(\frac{1}{b} \frac{\left(a+b e^x\right)^{6+1}}{6+1}\) + C
= \(\frac{1}{7 b}\) (a + be^x)⁷ + C

(ii) Let I = ∫ \(\frac{1}{x}\) (log x)² dx
put log x = t
⇒ d (log x) = dt
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ t² dt
= \(\frac{t^{3}}{3}\) + c
= \(\frac{(\log x)^3}{3}\) + c

Question 5.
(i) ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
(ii) ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
= ∫ (3 + log x)^1/2 \(\frac{1}{x}\) dx
[∵ \(\frac{d}{d x}\) (3 + log x) = 0 + \(\frac{1}{x}\) = \(\frac{1}{x}\)]
= \(\frac{(3+\log x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (3 + log x)^3/2 + C

(ii) Let I = ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
= 3 ∫ (x + log x)² (1 + \(\frac{1}{x}\)) dx
[∵ \(\frac{d}{d x}\) (x + log x) = 1 + \(\frac{1}{x}\)]
= 3 \(\frac{(x+\log x)^3}{3}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= (x + log x)³ + C

Question 6.
(i) ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
(ii) ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
= ∫ \(\left(\tan ^{-1} x\right)^{\frac{1}{2}} \cdot \frac{1}{1+x^2}\) dx
= \(\frac{\left(\tan ^{-1} x\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ \(\frac{d}{d x}\) tan^-1 x = \(\frac{1}{1+x^2}\)]
= \(\frac{2}{3}\left(\tan ^{-1} x\right)^{\frac{3}{2}}\) + C

(ii) As \(\frac{d}{d x}\) sin^-1 2x = \(\frac{1}{\sqrt{1-4 x^2}}\) × 2
∴ ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx = \(\frac{1}{2}\) ∫ (sin^-1 2x)³ \(\frac{2}{\sqrt{1-4 x^2}}\) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{1}{2} \frac{\left(\sin ^{-1} 2 x\right)^{3+1}}{(3+1)}\) + C
= \(\frac{1}{8}\) (sin^-1 2x)⁴ + C

Question 7.
(i) ∫ sin x cos⁵ x dx
(ii) ∫ cos³ x (ax + b) sin (ax + b) dx
Solution:
(i) Let I =∫ sin x cos⁵ x dx
= – ∫ (cos x)⁵ (- sin x) dx
= – \(\frac{(\cos x)^{5+1}}{5+1}\) + C
= – \(\frac{1}{6}\) cos⁶ x + C

(ii) Let I = ∫ cos³ x (ax + b) sin (ax + b) dx
= – \(\frac{1}{a}\) ∫ (cos (ax + b))³ (- a sin (ax + b)) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= – \(\frac{1}{a} \frac{(\cos (a x+b))^4}{4}\) + C
= – \(\frac{1}{4 a}\) cos⁴ (ax + b) + C

Question 8.
(i) ∫ sec³ x tan x dx
(ii) ∫ cot³ x cosec² x dx
Solution:
(i) Let I = ∫ sec³ x tan x dx
= ∫ sec² x (sec x tan x) dx
= \(\frac{(\sec x)^{2+1}}{2+1}\) + C
= \(\frac{\sec ^3 x}{3}\) + C

(ii) Let I = ∫ cot³ x cosec² x dx
= ∫ (cot x)³ cosec² x dx
put cos x = t
⇒ d (cot x) = dt
⇒ – cosec² x dx = dt
⇒ cosec² x dx = – dt
∴ I = ∫ t³ (- dt)
= \(\frac{-t^4}{4}\) + c
= \(\frac{- 1}{4}\) cot⁴ x + c

Question 9.
(i) ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
(ii) ∫ \(\frac{\sin x}{1+\cos x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
= ∫ (tan x)^{\(\frac{1}{2}\)} sec² x dx
= \(\frac{(\tan x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{2}{3} \tan \frac{3}{2} x\) + C

(ii) Let I = ∫ \(\frac{\sin x}{1+\cos x}\) dx
= – ∫ \(\frac{-\sin x d x}{1+\cos x}\)
= – log (1 + cos x) + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x) + C]

Question 10.
(i) ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
(ii) ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
= ∫ \(\frac{\cos x}{(\sin x)^{3 / 2}}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{d t}{t^{3 / 2}}\)
= ∫ t^-3/2 dt
= \(\frac{t^{-3 / 2+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= – 2t^-1/2 + c
= \(\frac{-2}{\sqrt{\sin x}}\) + c

(ii) Let I = ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
= ∫ \(\frac{\sin x}{(\cos x)^{3 / 2}}\) dx
put cos x = t
⇒ d (cos x) = dt
⇒ – sin x dx = dt
⇒ sin x dx = – dt
∴ I = ∫ \(\frac{-d t}{t^{3 / 2}}\)
= – \(\frac{t^{\frac{-3}{2}+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= \(\frac{2}{t^{1 / 2}}\) + c
= \(\frac{2}{\sqrt{\cos x}}\) + c

Question 11.
(i) ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
(ii) ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx.
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
= – \(\frac{1}{2}\) ∫ (3 + 2 cos x)^{– \(\frac{1}{2}\)} (- 2 sin x) dx
= – \(\frac{1}{2} \frac{(3+2 \cos x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= – \(\sqrt{3+2 cos x}\) + C

(ii) Let I = ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx
put x – cos x = t
⇒ (1 + sin x) dx = dt
∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)
= ∫ t^-1/2 dt
= \(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\) + c
= 2√t + c
= 2 \(\sqrt{x-cos x}\) + c

Question 12.
(i) ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
(ii) ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
= ∫ (1 + cos x)^-2 sin x dx
= – ∫ (1 + cos x)^-2 (- sin x dx)
= – \(\frac{(1+\cos x)^{-2+1}}{-2+1}\) + C
= \(\frac{1}{1+\cos x}\) + C

(ii) Let I = ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Since \(\frac{d}{d x}\) (tan³ x + 2 tan x + 5)
= 3 tan² x sec² x + 2 sec² x
= sec² x (3 tan² x + 2)
∴ I = ∫ (tan³ x + 2 tan x + 5)^{– 2} {(3 tan² x + 2) sec² x dx}
= \(\frac{\left(\tan ^3 x+2 \tan x+5\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{\tan ^3 x+2 \tan x+5}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]

Question 13.
(i) ∫ (e^3x + 1)² e^3x dx
(ii) ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
Solution:
(i) Let I = ∫ (e^3x + 1)² e^3x dx
[∵ \(\frac{d}{d x}\) (e^3x + 1) = 3e^3x]
= \(\frac{1}{3}\) ∫ (e^3x + 1)² (3e^3x dx)
= \(\frac{1}{3} \frac{\left(e^{3 x}+1\right)^3}{3}\) + C
= \(\frac{1}{9}\) (e^3x + 1)³ + C

(ii) Let I = ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
= ∫ (1 + e^4x)^{– 3} e^4x dx
= \(\frac{1}{4}\) ∫ (1 + e^4x)^{– 2} 4 e^4x dx
= \(\frac{1}{4} \frac{\left(1+e^{4 x}\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{4} \frac{1}{\left(1+e^{4 x}\right)}\) + C

Question 14.
(i) ∫ tan⁴ x dx
(ii) ∫ cot⁴ x dx
Solution:
(i) Let I = ∫ tan⁴ x dx
= ∫ tan² x . tan² x dx
= ∫ tan² x (sec² x – 1)
= ∫ tan² x sec² x dx – ∫ tan² x dx
= ∫ (tan x)² sec² x dx – ∫ (sec² x – 1) dx
= \(\frac{(\tan x)^{2+1}}{(2+1)}\) – tan x + x + C
= \(\frac{\tan ^3 x}{3}\) – tan x + x + C

(ii) Let I = ∫ cot⁴ x dx
= ∫ cot² x . cot² x dx
= ∫ cot² x (cosec² x – 1) dx
= ∫ (cot x)² cosec² x dx – ∫ (cosec² x – 1) dx
= – ∫ (cot x)² (- cosec² x dx) – ∫ cosec² x dx + ∫ dx + C
= – \(\frac{\cot ^3 x}{3}\) – cot x + x + C

Question 15.
(i) ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx
(ii) ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x + log sec x) = 1 + \(\frac{1}{\sec x}\) × sec x tan x
= 1 + tan x
∴ ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx = log |x + log sec x| + C
[∵∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Since \(\frac{d}{d x}\) (x + sin² x) = 1 + 2 sin x cos x
= 1 + sin 2x
∴ I = ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
= log |x + sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 16.
(i) ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
(ii) ∫ \(\frac{x^5}{1+x^6}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (1 – tan² x) = – 2 tan x sec² x
Let I = ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
= – \(\frac{1}{2} \int \frac{-2 \tan x \sec ^2 x d x}{1-\tan ^2 x}\)
= – \(\frac{1}{2}\) log |1 – tan² x| + C

(ii) Since \(\frac{d}{d x}\) (1 + x⁶) = 6x⁵
I = ∫ \(\frac{x^5}{1+x^6}\) dx
= ∫ \(\frac{6 x^5 d x}{1+x^6}\)
= \(\frac{1}{6}\) log |1 + x⁶| + C

Question 17.
(i) ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
= \(\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{3 \cos x+2 \sin x}\) dx
put 3 cos x + 2 sin x = t
⇒ d (3 cos x + 2 sin x) = dt
⇒ (- 3 sin x + 2 cos x) dx = dt
⇒ (2 cos x – 3 sin x) dx = dt
∴ I = ∫ \(\frac{d t}{2 t}\)
= \(\frac{1}{2}\) log |t| + c
= \(\frac{1}{2}\) log |3 cos x + 2 sin x| + c

(ii) Since \(\frac{d}{d x}\) (sin² x + cos x + 7) = 2 sin x cos x – sin x
= sin 2x – sin x
∴ I = ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
= log |sin² x + cos x + 7| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 18.
(i) ∫ \(\frac{x+\tan x}{x^2-2 \log \cos x}\) dx
(ii) ∫ \(\frac{1}{x \log x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x² – 2 lo0g cos x) = 2x – \(\frac{2}{\cos x}\) (- sin x)
= 2x + 2 tan x
Let I = ∫ \(\frac{(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) ∫ \(\frac{2(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) log |x² – 2 log cos x| + C

(ii) Let I = ∫ \(\frac{d x}{x \log x}\) ;
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |log x| + c

Question 19.
(i) ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
(ii) ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
= \(\frac{\frac{2 x^2-1}{x}}{x^2-\log x}\) dx
= \(\frac{\left(2 x-\frac{1}{x}\right) d x}{x^2-\log x}\) dx
As \(\frac{d}{d x}\) (x² – log x) = 2x – \(\frac{1}{x}\)
I = log |x² – log x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Divivde Num and Deno. by e^x ; we have
= ∫ \(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
put e^x – e^{– x} = t
⇒ (e^x + e^{– x}) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |e^x – e^{– x}| + C.

Question 20.
(i) ∫ \(\frac{1}{1+e^{-x}}\) dx
(ii) ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1}{1+e^{-x}}\) dx
= ∫ \(\frac{e^x d x}{1+e^x}\)
= log |1 + e^x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
= ∫ \(\frac{\frac{\sin x d x}{\cos x}}{a+b \frac{\sin ^2 x}{\cos ^2 x}}\)
= ∫ \(\frac{\sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
Since, \(\frac{d}{d x}\) (a cos² x + b sin² x)
= 2a cos x (- sin x) + 2b sin x cos x
= 2 (b – a) cos x sin x
∴ I = \(\frac{1}{2(b-a)} \int \frac{2(b-a) \sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
= \(\frac{1}{2(b-a)}\) log |a cos² x + b sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 21.
(i) ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
(ii) ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
= ∫ \(\frac{1-\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}\) dx
= ∫ \(\frac{(\sin x-\cos x)}{\sin x+\cos x}\) dx
putting sin x + cos x = t
⇒ d (sin x + cos x) = dt
⇒ (cos x – sin x) dx = dt
⇒ (sin x – cos x) dx = – dt
∴ I = ∫ \(-\frac{d t}{t}\)
= – log |t| + c
= – log |cos x + sin x| + c

(ii) Let I = ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Thus I = ∫ \(\frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2}\) dx
= ∫ \(\frac{\cos x-\sin x}{\cos x+\sin x}\) dx
put cos x + sin x = t
⇒ d (cos x + sin x) = dt
⇒ (- sin x + cos x) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |sin x + cos x| + c

After:
I = ∫ \(\frac{\cos 2 x d x}{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}\)
= \(\frac{\cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{2 \cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{1}{2}\) log |1 + sin 2x| + c
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]