Question

The value of \(\int \frac{\sin x + \cos x}{3 + \sin 2x} dx\) is

• A. \(\frac{1}{4} \log \left( \frac{2 - \sin x + \cos x}{2 + \sin x - \cos x} \right) + C\)
• B. \(\frac{1}{2} \log \left( \frac{2 + \sin x}{2 - \sin x} \right) + C\)
• C. \(\frac{1}{4} \log \left( \frac{1 + \sin x}{1 - \sin x} \right) + C\)
• D. None of the above

Hint:

Converting \(\sin 2x\) to expression in terms of \(\sin x - \cos x\) simplifies the integral.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Substitute \(t = \sin x - \cos x\) or \(t = \sin x + \cos x\).

Step 2: Detailed Explanation:
Note that \(3 + \sin 2x = 3 + 2\sin x \cos x\). Also \((\sin x - \cos x)^2 = 1 - \sin 2x\), so \(\sin 2x = 1 - (\sin x - \cos x)^2\). Let \(t = \sin x - \cos x\). Then \(dt = (\cos x + \sin x) dx\). Also \(3 + \sin 2x = 3 + 1 - t^2 = 4 - t^2\). Integral becomes \(\int \frac{dt}{4 - t^2} = \frac{1}{4} \log \left| \frac{2+t}{2-t} \right| + C\). Substitute back: \(t = \sin x - \cos x\). Then \(\frac{2+t}{2-t} = \frac{2 + \sin x - \cos x}{2 - \sin x + \cos x}\). But the given option has \(\frac{2 - \sin x + \cos x}{2 + \sin x - \cos x}\) which is the reciprocal. The log of reciprocal gives a negative sign. So the integral is \(-\frac{1}{4} \log \left( \frac{2 + \sin x - \cos x}{2 - \sin x + \cos x} \right) + C = \frac{1}{4} \log \left( \frac{2 - \sin x + \cos x}{2 + \sin x - \cos x} \right) + C\). Yes, matches.

Step 3: Final Answer:
Option (A) is correct.