Step 1: Plan:
Use the half-angle substitution style to get $I_c$ directly in terms of $I_1$.
Step 2: Steps:
Look at the difference $I_c - I_s = \int\frac{\cos x-\sin x}{1+\sin x+\cos x}dx$.
The numerator is the derivative of $1+\sin x+\cos x$, so this is $\left[\log(1+\sin x+\cos x)\right]$ from $\frac{\pi}{6}$ to $\frac{\pi}{3}$. The two end values are $1+\frac{\sqrt3}{2}+\frac12$ at both limits, so the difference is $0$. So $I_c = I_s$.
Also $I_c + I_s = \frac{\pi}{6} - \log2$, using $\int dx = \frac{\pi}{6}$ and the given integral. Therefore $I_c = \frac12\left(\frac{\pi}{6}-\log2\right) = \frac{\pi}{12}-\frac12\log2$.
Final Answer:
The integral equals $\frac{\pi}{12}-\frac12\log2$, option (A).
\[ \boxed{\frac{\pi}{12}-\frac{1}{2}\log2} \]