Evaluate: $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ 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?

$$I=\displaystyle \int_0^{\pi} \ln \left( \sin \theta \right)\cdot  d\theta$$
$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$
$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$
Adding both.
$$I=\displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \times \cos \theta\right) \cdot 
d\theta$$
$$I= \displaystyle \int_0^{\pi/2} \ln \left(2 \sin \theta\times \cos \theta \right) -\ln2 \cdot d\theta$$
$$I=\int_0^{\pi/2}\ln(\sin{2\theta})-\ln2 \cdot d\theta$$
$$\int_0^{\pi/2}\ln(\sin{2\theta})\cdot d\theta=I/2$$
So,
$$I=-2\int_0^{\pi/2}\ln2\cdot d\theta$$
$$I=-{\pi\ln2}$$
