Question

\(\int\frac{2\cos 2x}{(1+\sin 2x)(1+\cos 2x)}dx=\)

• A. \(2\tan x + \log(1+\tan x) + c\)
• B. \(\tan x - 2\log(1+\tan x) + c\)
• C. \(2\log(1+\tan x) + \tan x + c\)
• D. \(2\log(1+\tan x) - \tan x + c\)

Hint:

Use trigonometric identities to simplify the integrand and make appropriate substitutions.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand.
We have \(\sin 2x = 2\sin x \cos x\) and \(\cos 2x = 2\cos^2 x - 1 = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x\).
Also, \(1 + \cos 2x = 2\cos^2 x\) and \(1 + \sin 2x = (\sin x + \cos x)^2\).
The integral becomes: \(\int \frac{2\cos 2x}{(1+\sin 2x)(1+\cos 2x)}dx = \int \frac{2(\cos^2 x - \sin^2 x)}{(\sin x + \cos x)^2 (2\cos^2 x)} dx\)
= \(\int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2 \cos^2 x} dx\)
= \(\int \frac{\cos x - \sin x}{(\cos x + \sin x) \cos^2 x} dx\)
= \(\int \frac{\cos x - \sin x}{\cos^2 x (1 + \tan x)} dx\)
= \(\int \frac{\frac{\cos x}{\cos^2 x} - \frac{\sin x}{\cos^2 x}}{1 + \tan x} dx\)
= \(\int \frac{\sec x - \sec x \tan x}{1 + \tan x} dx\)
= \(\int \frac{\sec x (1 - \tan x)}{1 + \tan x} dx\)

Step 2: Use the substitution t = \(\tan x\).
Then dt = \(\sec^2 x dx\).
We need to rewrite the integral in terms of \(\tan x\).
\(\int \frac{\cos x - \sin x}{(\cos x + \sin x) \cos^2 x} dx = \int \frac{\frac{\cos x - \sin x}{\cos^2 x}}{\frac{\cos x + \sin x}{\cos^2 x}} dx\)
= \(\int \frac{\sec x - \tan x \sec x}{1 + \tan x} dx\)
= \(\int \frac{1 - \tan x}{1 + \tan x} \sec^2 x \frac{1}{\sec x} dx\)
= \(\int \frac{1 - \tan x}{1 + \tan x} \frac{1}{\cos x} dx\)

Step 3: Use the substitution u = 1 + \(\tan x\).
Then du = \(\sec^2 x dx\).
We have \(\cos 2x = \cos^2 x - \sin^2 x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} = \frac{1 - \tan^2 x}{1 + \tan^2 x}\).
\(\sin 2x = 2\sin x \cos x = \frac{2\sin x \cos x}{\cos^2 x + \sin^2 x} = \frac{2\tan x}{1 + \tan^2 x}\).
\(\int \frac{2\cos 2x}{(1+\sin 2x)(1+\cos 2x)} dx = \int \frac{2(1 - \tan^2 x)}{(1 + 2\tan x + \tan^2 x)(2\cos^2 x)} dx\)
= \(\int \frac{1 - \tan^2 x}{(1 + \tan x)^2 \cos^2 x} dx\)
= \(\int \frac{(1 - \tan x)(1 + \tan x)}{(1 + \tan x)^2 \cos^2 x} dx\)
= \(\int \frac{1 - \tan x}{1 + \tan x} \sec^2 x dx\)
Let u = 1 + \(\tan x\). Then du = \(\sec^2 x dx\).
\(\tan x = u - 1\).
\(\int \frac{1 - (u - 1)}{u} du = \int \frac{2 - u}{u} du\)
= \(\int (\frac{2}{u} - 1) du = 2\log|u| - u + c\)
= \(2\log|1 + \tan x| - (1 + \tan x) + c\)
= \(2\log|1 + \tan x| - \tan x - 1 + c\)
= \(2\log|1 + \tan x| - \tan x + c'\)
Therefore, the integral is \(2\log(1+\tan x) - \tan x + c\).