Question:

\( \int \frac{1}{\sqrt{1 + \cos 2x}} \, dx \) is equal to :

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Always look for ways to eliminate square roots in denominators using double angle identities (\( \cos 2x \)).
Remember: \( 1 - \cos 2x = 2\sin^2 x \) and \( 1 + \cos 2x = 2\cos^2 x \).
Updated On: Sep 10, 2026
  • \( \log \cos x + C \)
  • \( \frac{1}{\sqrt{2}} \log |\sec x + \tan x| + C \)
  • \( \frac{1}{\sqrt{2}} \log |\sec x - \tan x| + C \)
  • \( \log \sin 2x + C \)
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The Correct Option is B

Solution and Explanation

Concept:
• Use trigonometric identity: \( 1 + \cos 2x = 2\cos^2 x \).
• Standard Integral: \( \int \sec x \, dx = \log |\sec x + \tan x| + C \).

Step 1:
Simplify the integrand using trig identities
Let \( I = \int \frac{1}{\sqrt{1 + \cos 2x}} \, dx \). Substitute \( 1 + \cos 2x = 2\cos^2 x \): \[ I = \int \frac{1}{\sqrt{2\cos^2 x}} \, dx \] \[ I = \int \frac{1}{\sqrt{2} \cos x} \, dx \]

Step 2:
Transform the integral into a standard form
Factor out the constant \( 1/\sqrt{2} \): \[ I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos x} \, dx \] \[ I = \frac{1}{\sqrt{2}} \int \sec x \, dx \]

Step 3:
Integrate and add the constant of integration
Applying the standard formula for \( \int \sec x \, dx \): \[ I = \frac{1}{\sqrt{2}} \log |\sec x + \tan x| + C \] This result matches option (B).
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