Verify the following equation:
$$ \int_{0}^{2\pi}{\log{\left(\sin^{2}{2\theta}\right)}\,\mathrm{d}\theta} = 4\int_{0}^{\pi}{\log{\left(\sin{\theta}\right)}\,\mathrm{d}\theta} = -4\pi \log{2}$$
I am unable to think of a closed rectifiable curve so that I can apply the residue theorem to prove the above equation. At first I thought we can put $z=e^{i\theta}$ and take $0\leq \theta \leq 2\pi $ but then I could not understand the singularities and analyticity of the log in that.
I did not find a way to perform the integration by evaluating a residue at a singularity of a function. However I will present below a hint how to calculate the integral using contour integration of an analytic function.
$$\begin{align} \int_0^{\pi} \log(\sin x)\,dx &=\int_0^{\pi} \log\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\,dx\\ &=\int_0^{\pi} \left[\log\left(1-e^{-2ix}\right)+ix-\log(2i)\right]dx\\ &=\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx+\frac{i\pi^2}2-\pi\log2-\frac{i\pi^2}2\\ &=\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx-\pi\log2, \end{align}$$ where we used $\log(i)=\frac{i\pi}2$.
So we have to prove $\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx=0$. This can be done by the integration over the following path: $$\begin{array}{lllll} (1):&\quad \epsilon&\to&\pi-\epsilon\\ (2):&\quad \pi-\epsilon&\to& \pi+i \epsilon\\ (3):&\quad \pi+i\epsilon&\to& \pi+iR\\ (4):&\quad \pi+iR &\to& iR\\ (5):&\quad iR&\to& i\epsilon\\ (6):&\quad i\epsilon&\to&\epsilon\\ \end{array}$$ Observe that the integrals along $(3)$ and $(5)$ cancel each other, hence it remains to show that the integrals along $(2),(4),(6)$ tend to $0$ as $\epsilon\to0,\; R\to\infty$, which is left as an exercise.