Is it possible to evaluate this integral? Is it possible to evaluate this integral: $$\int_{0}^{\frac{\pi }{2}}\ln(\sin 2x){\rm d}x$$
 A: Yes, it is possible. Here is how you advance. Recalling the identity $\sin(2x)=2\sin(x)\cos(x)$, we have
$$ \int_{0}^{\pi/2}\ln(\sin(2x))dx =\int_{0}^{\pi/2}\ln(2)dx + \int_{0}^{\pi/2}\ln(\sin(x))dx + \int_{0}^{\pi/2}\ln(\cos(x))dx. $$
The first integral is easy to evaluate. The second and the third integrals are equal and you can see this by just using the change of variables $x=\frac{\pi}{2}-\theta$. Now, let's evaluate one of them as
$$ \int_{0}^{\pi/2}\ln(\sin(x))dx = \int_{0}^{1}\frac{\ln(u)}{\sqrt{1-u^2}}du=\frac{1}{4}\int_{0}^{1}\ln(t)t^{-1/2}(1-t)^{-1/2}dt=I, $$
where the changes of variables $\sin(x)=u$ and $u^2=t$ were used respectively. To evaluate the last integral, consider the integral
$$ F = \int_{0}^{1}t^{\alpha}(1-t)^{\beta}dt = \beta(\alpha+1,\beta+1), $$
where $\beta(s,t)$ is the $\beta$ function. So, $I$ follows from $F$ as
$$ I= \frac{1}{2}\lim_{\beta \to -1/2}\lim_{\alpha \to -1/2}F_{\alpha}(\alpha,\beta)=-\frac{1}{2}\pi\ln(2). $$
A: It is well known that  $\Pi_{1\le k<n}\sin \frac{k\pi}{n}=\frac{n}{2^{n-1}}$.
So $\int_0^\pi \ln \sin x dx =\lim_{n\to \infty}\frac{\pi}{n}\ln \Pi_{1\le k<n} \sin \frac{k\pi}{n}=-\pi\ln 2 $. Therefore $\int_0^{\pi/2} \ln \sin 2x dx=-\pi\frac{\ln 2}{2}$.
A: This link contains at least four different ways to compute it (none of them uses Clausen function).
