How to compute $\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$ I am trying to compute this integral.
$$\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$$
Any thoughts will help. Thanks.
 A: Note that
\begin{eqnarray}
&&\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x\\
&=&\int_0^{\frac{\pi}{2}}(\cos x)^{-1}\sin x\ln(\sin x)\ \text{d}x=\lim_{a\to0,b\to2}\frac{d}{db}\int_0^{\frac{\pi}{2}}(\cos x)^{a-1}(\sin x)^{b-1}\ \text{d}x\\
&=&\lim_{a\to0,b\to2}\frac{d}{db}\frac{1}{2}B\left(\frac{a}{2},\frac{b}{2}\right)\\
&=&\lim_{a\to0,b\to2}\frac{1}{4}B\left(\frac{a}{2},\frac{b}{2}\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+b}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi(1)-\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\frac{\Gamma(\frac{a}{2})\Gamma(1)}{\Gamma(\frac{a+2}{2})}\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\left(\frac{2}{a}+O(a^2)\right)\left(\frac{\pi^2}{12}a+O(a^2)\right)\\
&=&-\frac{\pi^2}{24}.
\end{eqnarray}
Here we use the following fact
$$ \Gamma(a)\approx\frac{1}{a}, \gamma+\psi(\frac{a}{2}+1)\approx\frac{\pi^2}{12}a.$$
A: A partial answer: Let $I$ be the given integral. You can integrate by parts, letting
$$\begin{matrix}u=\dfrac{\ln(\sin x)}{\cot x}&&&\mathrm{d}v=\mathrm{d}x\\[1ex]
\mathrm{d}u=\left(1+\sec^2x\ln(\sin x)\right)\,\mathrm{d}x&&&v=x\end{matrix}$$
which yields
$$\begin{align*}
I&=\frac{x\ln(\sin x)}{\cot x}\Bigg|_{x=0}^{x=\pi/2}-\int_0^{\pi/2} x\left(1+\sec^2x\ln(\sin x)\right)\,\mathrm{d}x\\[1ex]
&=0-\left(\int_0^{\pi/2}x\,\mathrm{d}x+\int_0^{\pi/2} x\sec^2x\ln(\sin x)\,\mathrm{d}x\right)\\[1ex]
&=-\frac{\pi^2}{8}-\int_0^{\pi/2} x\sec^2x\ln(\sin x)\,\mathrm{d}x
\end{align*}$$
It turns out that the remaining integral is twice $I$, which means $3I=-\dfrac{\pi^2}{8}$, or $I=-\dfrac{\pi^2}{24}$. I'm still seeing what can be done to establish this last part, though.
A: Note that 
$$\int_{0}^{\frac{\pi}{2}}{\frac{\ln(\sin{x})}{\cot{x}}}\,dx = \int_{0}^{\frac{\pi}{2}}{\ln(\sin{x})\tan{x}}\,dx = \int_{0}^{\frac{\pi}{2}}\frac{\ln(\sin{x})\sin{x}}{\cos{x}}\,dx $$
$$= \int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\sqrt{1-\cos^2{x}}\right)\sin{x}}{\cos{x}}\,dx$$
Now, set $u=\cos{x}\implies du=-\sin{x}\,dx$. Also, note that $\ln\left(\sqrt{1-u^2}\right)=\frac{1}{2}\ln\left(1-u^2\right)$ Also, don’t forget to change the limits. Now, we have
$$\int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\sqrt{1-\cos^2{x}}\right)\sin{x}}{\cos{x}}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\ln\left(1-u^2\right)}{u}\,du$$
There are two paths you can take from here. One I have shown below, and the other is shown by Dr. MV in the comments below. 
Now, we need to find what $\ln\left(1-u^2\right)$ is. Recall the Maclaurin Series for a natural logarithm. Therefore,
$$\ln\left(1-u^2\right)=-\sum_{n=1}^{\infty}\frac{u^{2n}}{n}$$
$$\therefore \frac{1}{2}\int_{0}^{1}\frac{\ln\left(1-u^2\right)}{u}\,du=-\frac{1}{2}\int_{0}^{1}\sum_{n=1}^{\infty}\frac{u^{2n}}{nu}\,du=-\frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{1}\frac{u^{2n-1}}{n}\,du$$
$$=-\frac{1}{4}\sum_{n=1}^{\infty}\left.\frac{u^{2n}}{n^2}\right|_{u=0}^{u=1}=-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$
Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$,
$$\int_{0}^{\frac{\pi}{2}}{\frac{\ln(\sin{x})}{\cot{x}}}\,dx=-\frac{\pi^2}{24}$$
A: Hint: $\cot(x) = (\ln(\sin(x)))'$
