Question:
The integral \( \int \sqrt{1 + \sin 2x} \, dx \) is equal to:
The integral \( \int \sqrt{1 + \sin 2x} \, dx \) is equal to:
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Trigonometric identities help simplify integrals involving trigonometric functions.
Updated On: Mar 7, 2025
- \( \sin x - \cos x + C \)
- \( \sin x - \csc x + C \)
- \( \tan x - \cot x + C \)
- \( \cos x - \sec x + C \)
- \( \tan x - \sec x + C \)
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The Correct Option is A
Solution and Explanation
Step 1:
Simplify the integrand using the trigonometric identity for \( \sin 2x \):
\[
\int \sqrt{1 + \sin 2x} \, dx = \int \left( \sin x - \cos x + C \right) \, dx.
\]
Step 2:
This integral can now be solved directly:
\[
\int \left( \sin x - \cos x \right) \, dx = \sin x - \cos x + C.
\]
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