Question

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

• A. \( \sin x - \cos x + k \)
• B. \( - \sin x - \cos x + k \)
• C. \( \sin x + \cos x + k \)
• D. \( - \sin x + \cos x + k \)

Hint:

When dealing with integrals involving trigonometric identities, it helps to express functions using standard trigonometric identities to simplify the problem.

Answer 1

The Correct Option is C

We are given: \[ I = \int \frac{\cos 2x}{\cos x + \sin x} dx \] To solve this, first, recall the double angle identity for cosine: \[ \cos 2x = 2 \cos^2 x - 1 \] Substitute this into the integral: \[ I = \int \frac{2 \cos^2 x - 1}{\cos x + \sin x} dx \] Now use substitution to simplify the integral. Let \( u = \cos x + \sin x \), then \( du = (\cos x - \sin x) dx \). This helps to simplify the integral and can lead us to the solution after applying basic integration techniques. The final answer, after proper computation, is: \[ I = \sin x + \cos x + k \]