TITLE: Evaluate: $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$
QUESTION [23 upvotes]: Evaluate: $ \displaystyle \int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ using Gauss Mean Value theorem.
Given hint: consider $f(z) = \ln ( 1 +z)$.
EDIT:: I know how to evaluate it, but I am looking if I can evaluate it using Gauss MVT.
ADDED:: Here is what I have got so far!!
$$\ln 2 = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{i \theta}) d\theta = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{-i \theta}) d\theta$$
Hence, $ \displaystyle 2 \ln 2 = \frac{1}{2 \pi } \int_{0}^{2 \pi} \log(5 + 4 \cos \theta )d \theta  = \frac{1}{\pi} \int_0^{\pi} \log(1 + 8 \cos^2 \theta) d \theta$,  now to problem is how to reduce it to the above form?

REPLY [11 votes]: This is just for fun. 
It is well known that  $\Pi_{1\le k<n}\sin \frac{k\pi}{n}=\frac{n}{2^{n-1}}$. So $\int_0^\pi \ln \sin x dx =\lim_{n\to \infty}\frac{\pi}{n}\ln \Pi_{1\le k<n} \sin \frac{k\pi}{n}=-\pi\ln 2 $.