Showing $\int_0^{\pi} \log(2 - 2 \cos x) = 0$. I want to show that
$$\int_0^{\pi} \log(2 - 2 \cos x) = 0$$
However, I cannot do this.  I tried splitting the integral into $\int_0^{\pi/3} \log(2 - 2 \cos x)\,dx + \int_{\pi/3}^{\pi} \log(2 - 2 \cos x) \,dx$ and showing that the two parts were negatives of one another.  Wolframalpha does not give very a simple antiderivative.  I was wondering if there was a nice way to do this. 
Other attempts: using $\int_0^a f(x) \,dx = \int_0^{a} f(a-x) \,dx$, trying to change the $\cos$ to $\sin$ by some substitution like $u = \pi/2 - x$ and trying to get things to cancel. 
 A: First we prove

$$\displaystyle\int_0^{\pi/2} \log  \left(\sin x\right)\,dx =-\frac12\pi\log2$$

Proof $\ $
From the substitution $x \to \frac{\pi}{2}-x$ , we get
$$
\displaystyle\int_0^{\pi/2} \log( \sin x  )\,dx
=
\displaystyle\int_0^{\pi/2} \log ( \cos x  )\,dx 
$$
Thus
\begin{align}
2\displaystyle\int_0^{\pi/2} \log( \sin x  )\,dx
 &=
\displaystyle\int_0^{\pi/2} \log (\sin x \cos x )\,dx
\\ &=
\displaystyle\int_0^{\pi/2} \log (\sin 2x )\,dx-\frac{1}{2}\pi\log 2 \\
&=\frac12\displaystyle\int_0^{\pi} \log (\sin x )\,dx-\frac{1}{2}\pi\log 2 \\
&= \displaystyle\int_0^{\pi/2} \log( \sin x  )\,dx -\frac{1}{2}\pi\log 2
\end{align}
Then we arrive at the conclusion. 
To calculate the original integral, we have
\begin{align}
\int_0^\pi \log(2-2\cos x)\,dx
&= \int_0^\pi \log(4\sin^2 \frac{x}{2})\,dx \\
&= 2\pi \log 2 + 2 \int_0^\pi \log (\sin \frac{x}{2}) \,dx \\
&= 2\pi \log 2 + 4\int_0^{\pi/2} \log  \left(\sin x\right)\,dx  \\
&= 0
\end{align}
A: $$\log(2-2\cos x)=\log(2(1-\cos x))=\log2+\log(2\sin^2 x/2)=2\log2+2\log\sin x/2$$
If we show that $\int_0^\pi\log\sin(x/2)~\mathrm dx=-\pi\log2$, we are done.
With $t\mapsto x/2$, the above claim is,

$$\int_0^{\pi/2}\log\sin t~\mathrm dt=-\frac \pi2\log2$$

This is a pretty well known integral. Can you take it from here?
A: $$\int_0^\pi  \log \left(4\sin^2\left(\frac{x}{2}\right)\right) dx$$ is equivalent.
Use logarithm rules to make it -
$$\int_0^\pi \left[ \log 4 + 2 \log\left(\sin\left(\frac{x}{2}\right)\right) \right] dx$$
The rightmost part of the integral has already been solved, which is linked here.
