Question

If \(\int _{π/6}^{π/3}\frac{1}{1+sinx+cosx}dx = log2\), then the value of \(\int _{π/6}^{π/3}\frac{cosx}{1+sinx+cosx}dx =\) ............

• A. \(\frac{π}{12}-\frac{1}{2}log2\)
• B. \(\frac{π}{12}-\frac{1}{3}log2\)
• C. \(\frac{π}{6}-\frac{1}{2}log2\)
• D. \(\frac{π}{12}-\frac{1}{4}log2\)

Hint:

The interval is symmetric under \(x\to\frac{\pi}{2}-x\), so the cosine and sine integrals are equal.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Let \(I_1 = \int\frac{1}{1+\sin x+\cos x}dx = \log2\), \(I_c = \int\frac{\cos x}{1+\sin x+\cos x}dx\) and \(I_s = \int\frac{\sin x}{1+\sin x+\cos x}dx\), all from \(\frac{\pi}{6}\) to \(\frac{\pi}{3}\).

Step 2: Add:
\[ I_c + I_s = \int\frac{\sin x+\cos x}{1+\sin x+\cos x}dx = \int\left(1 - \frac{1}{1+\sin x+\cos x}\right)dx \]
\[ I_c + I_s = \left(\frac{\pi}{3}-\frac{\pi}{6}\right) - \log2 = \frac{\pi}{6}-\log2 \]

Step 3: Use symmetry:
Replace \(x\) by \(\frac{\pi}{2} - x\). The limits \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) swap with each other, and \(\sin\) and \(\cos\) swap. So \(I_c = I_s\).
Hence \(2I_c = \frac{\pi}{6}-\log2\), so \(I_c = \frac{\pi}{12} - \frac12\log2\).

Final Answer:
The integral equals \(\frac{\pi}{12}-\frac12\log2\), option (A). \[ \boxed{\frac{\pi}{12}-\frac{1}{2}\log2} \]