Question:
The integral \( \int \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \, dx \) is equal to :
The integral \( \int \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \, dx \) is equal to :
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The double angle identities \( 1-\cos 2x = 2\sin^2 x \) and \( 1+\cos 2x = 2\cos^2 x \) are highly frequent in calculus. Memorize them to simplify trigonometric roots quickly.
Updated On: Jul 6, 2026
- \( \log |\sec^2 x| + C \)
- \( \log |\cos x| + C \)
- \( \log |1 + \cos 2x| + C \)
- \( \log |\sec x| + C \)
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The Correct Option is D
Solution and Explanation
Concept:
We utilize trigonometric double-angle identities to simplify the integrand:
\[
1 - \cos 2x = 2\sin^2 x \quad \text{and} \quad 1 + \cos 2x = 2\cos^2 x
\]
The standard integral of the tangent function is given by \( \int \tan x \, dx = \log |\sec x| + C \).
Step 1: Substitute double-angle identities into the integrand. Let's simplify the inner fraction: \[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\frac{\sin^2 x}{\cos^2 x}} = \sqrt{\tan^2 x} = \tan x \] (Assuming standard principal domains where \( \tan x > 0 \)).
Step 2: Integrate the simplified function. The original integral now reduces directly to: \[ I = \int \tan x \, dx \] We solve this using substitution by rewriting \( \tan x = \frac{\sin x}{\cos x} \): \[ I = \int \frac{\sin x}{\cos x} \, dx \] Let \( u = \cos x \), then \( du = -\sin x \, dx \implies \sin x \, dx = -du \): \[ I = \int \frac{-du}{u} = -\log|u| + C = -\log|\cos x| + C \] Using logarithmic power rules, \( -\log|\cos x| = \log|(\cos x)^{-1}| = \log\left|\frac{1}{\cos x}\right| = \log|\sec x| + C \).
Step 1: Substitute double-angle identities into the integrand. Let's simplify the inner fraction: \[ \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\frac{\sin^2 x}{\cos^2 x}} = \sqrt{\tan^2 x} = \tan x \] (Assuming standard principal domains where \( \tan x > 0 \)).
Step 2: Integrate the simplified function. The original integral now reduces directly to: \[ I = \int \tan x \, dx \] We solve this using substitution by rewriting \( \tan x = \frac{\sin x}{\cos x} \): \[ I = \int \frac{\sin x}{\cos x} \, dx \] Let \( u = \cos x \), then \( du = -\sin x \, dx \implies \sin x \, dx = -du \): \[ I = \int \frac{-du}{u} = -\log|u| + C = -\log|\cos x| + C \] Using logarithmic power rules, \( -\log|\cos x| = \log|(\cos x)^{-1}| = \log\left|\frac{1}{\cos x}\right| = \log|\sec x| + C \).
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