Integral of $\log(\sin(x))$ using contour integrals I know the integral is possible with a simple fourier series expansion of $-\log(\sin(x))$
But I am interested in complex analysis, so I want to try this.
$$I = \int_{0}^{\pi} \log(\sin(x)) dx$$
The substitution $x = \arcsin(t)$ first comes to mind. 
But that substitution isnt valid as, 
The upper and lower bounds would both be $0$ because $\sin(\pi) = \sin(0) = 0$
What is a workaround using inverse trig or some other way?
Inverse sine is good, because that gives us a denominator from which we will be able to find poles.
$$x = arctan(t)$$
Is also good, but it would hard to do.
Idea?

Another solution is as follows.
Let 
$$ I = \int_0^{\pi} \log(\sin x)\, dx. $$
First note by symmetry, 
$$ I = 2\int_0^{\pi/2} \log(\sin x)\, dx. $$
Now letting $x = 2u$, $dx = 2du$, and $\sin{2u} = 2\sin{u}\cos{u}$,
\begin{align}
2\int_0^{\pi/2} \log(\sin x)\, dx &= 4\int_0^{\pi/4} \log(\sin 2u)\, du\\
&= 4\int_0^{\pi/4} \log(2\sin{u}\cos{u})\, du\\
&= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_0^{\pi/4} \log(\cos u)\, du.
\end{align}
In the last integral, letting $z = \pi/2 - u$, gives $dz = -du$ and $\cos u = \sin(z)$ and hence
\begin{align}
I &= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du - 4\int_{\pi/2}^{\pi/4} \log(\sin z)\, dz\\
&= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_{\pi/4}^{\pi/2} \log(\sin z)\, dz\\
&= \pi \log{2} + 2I.
\end{align}
Solving for $I$ shows
$$ I = -\pi \log{2}. $$
