Calculate an integral using complex integration came across this one
$$\int_0^{\pi / 2} \ln (\sin x)\;dx$$
I wanted to find it using the residues, but, I don't thing they are isolated ones
 A: Let us write $I$ this integral.
Changing the variable $x$ into $y=\frac\pi2-x$, we get
$$I=\int_0^{\pi/2}\ln(\cos y)\,\mathrm dy$$
so we can add up the two integrals
$$\begin{split}2I&=\int_0^{\pi/2}\left[\ln(\sin x)+\ln(\cos x)\right]\mathrm dx=\int_0^{\pi/2}\ln\left(\frac12\sin 2x\right)\,\mathrm dx\\&=-\frac\pi2\ln 2+\int_0^{\pi/2}\ln(\sin 2x)\,\mathrm dx\end{split}\tag{1}$$
Let us change the variable in the last integral into $z=2x$ :
$$\begin{split}\int_0^{\pi/2}\ln(\sin 2x)\,\mathrm dx&=\frac12\int_0^\pi\ln(\sin z)\,\mathrm dz\\&=\frac12\int_0^{\pi/2}\ln(\sin x)\,\mathrm dx+\frac12\int_{\pi/2}^\pi\ln(\sin x)\,\mathrm dx\\&=\frac12I+\frac12J.\end{split}\tag{2}$$
Then we use the  change of variable $t=\pi-x$ to compute $J$ :
$$J=\int_{\pi/2}^\pi\ln(\sin x)\,\mathrm dx=\int_0^{\pi/2}\ln(\sin t)\,\mathrm dt=I.$$
As a conclusion, we obtain that $2I=-\frac\pi2\ln2+I$, hence the result
$I=-\frac\pi2\ln2$.
A: Strange but true, it is enough to use the definition of the Riemann integral through Riemann sums.
$$ I = \int_{0}^{\pi}\log\sin x\,dx = \lim_{n\to +\infty}\frac{\pi}{ n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}\tag{1} $$
but
$$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n} = \frac{2n}{2^n}\tag{2} $$
is a well-known identity, giving:

$$ \int_{0}^{\pi}\log\sin(x)\,dx = \color{red}{-\pi\log 2}.\tag{3}$$

A: I think there's an easier way: try differentiation by parts, set $\int_{0}^{\frac{\pi}{2}} 1 \cdot \ln (\sin x)dx = x \log \sin x |_{0}^{\frac{\pi}{2}} -\int_{0}^{\frac{\pi}{2}} \frac{x \cos x dx}{\sin x}$
