How to evaluate $\int_0^\pi(\ln(\sin(x)))^2 \, dx$ To find: $$\int_0^\pi(\ln(\sin(x)))^2 \, dx$$  Tried changing limits and adding, also tried using complex numbers but failed.
 A: write $\sin(x)^a=e^{a\log(\sin(x))}$
Then
$$
I=\partial_a^2 \left. \int_0^\pi dx\sin^a(x) \right|_{a=0}
$$
This integral can be evaluated in terms of Euler's Beta function and yields after simplification
$$
I=\sqrt{\pi }\partial_a^2 \left. \left(\frac{ \Gamma \left(\frac{a+1}{2}\right)}{\Gamma \left(\frac{a}{2}+1\right)}\right) \right|_{a=0}
$$
calculating the derivatives and taking the limit as $a\rightarrow 0$
is a little bit exhausting, but finally  yields an result (best use a CAS), 

$$
I=\frac{\pi ^3}{12}+\pi  \log ^2(2)
$$

Essentially the limit follows from the fact that:
$$
 \psi_0\left(\frac{1}{2}\right)=-\gamma-2\log(2),\quad
\psi_1\left(\frac{1}{2}\right)=\frac{\pi^2}{2},\quad \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}
$$
and 
$$
\partial_x \Gamma(x)=\Gamma(x) \psi_0(x), \quad \partial_x \psi_0(x)=\psi_1(x)
$$
where $ \psi_{0/1}(x)$ is the Di/Trigamma function and $\gamma$ is the Euler-Mascheroni constant. 
A: Maybe a faster way to recover tired's result is to apply Parseval's identity to the Fourier series of $\log\sin x$.
