<s>[INST] Integrate square of the log-sine integral: $\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$ $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)$ is an integral that is common.
But, how can we show $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx=\frac{{\pi}^{3}}{24}+\frac{\pi}{2}\ln^{2}(2)$?.
Does anyone have any ideas on how to approach $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$?.
Thank you very much. 
 [/INST] Perform the change of variable \begin{align}y=\dfrac{\pi}{2}-x\end{align} 
\begin{align}J&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin x\right) \,dx\\
&=\int_0^{\frac{\pi}{2}}\ln^2\left(\cos x\right)\,dx\\\end{align}
Consider the integrals,
\begin{align}A&=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan x\right)\,dx\\
B&=\int_0^{\frac{\pi}{2}}\ln^2(\sin x\cos x)\,dx\end{align}
observe that,
\begin{align}A+B=4J\end{align}
Perform the change of variable \begin{align}y=\tan x\end{align}
\begin{align}A&=\int_0^\infty \frac{\ln^2 x}{1+x^2}\,dx\end{align}
Consider the double integral
\begin{align} K&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=2\int_0^\infty \int_0^\infty \frac{\ln^2 x}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\pi A
\end{align}
since,
\begin{align} \int_0^\infty \frac{\ln x}{1+x^2}\,dx=0\end{align}
On the other hand, perform the change of variable $u=xy$,
\begin{align}K&=\int_0^\infty \int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}\,du\,dy\\\end{align}
Perform the change of variable $v=y^2$,
\begin{align}K&=\frac{1}{2}\int_0^\infty \int_0^\infty \frac{\ln^2 u}{(u^2+v)(1+v)}\,du\,dv\\
&=\frac{1}{2}\int_0^\infty\left[\frac{\ln\left(\frac{v+1}{v+u^2}\right)}{u^2-1}\right]_{v=0}^{\infty}\ln^2 u\,du\\
&=\int_0^\infty\frac{\ln^3 u}{u^2-1}\,du\\
&=\int_0^1\frac{\ln^3 u}{u^2-1}\,du+\int_1^\infty\frac{\ln^3 u}{u^2-1}\,du\\
\end{align}
In the latter integral perform the change of variable $z=\dfrac{1}{u}$,
\begin{align}K&=2\int_0^1\frac{\ln^3 u}{u^2-1}\,du\\
&=2\int_0^1\frac{\ln^3 u}{u-1}\,du-\int_0^1\frac{2u\ln^3 u}{u^2-1}\,du
\end{align}
In the latter integral perform the change of variable $z=u^2$,
\begin{align}K&=\left(2-\frac{1}{8}\right)\int_0^1\frac{\ln^3 u}{u-1}\,du\\
&=-\frac{15}{8}\int_0^1\left( \sum_{n=0}^\infty u^{n}\right)\ln^3 u\,du\\
&=-\frac{15}{8}\sum_{n=0}^\infty\int_0^1 u^{n}\ln^3 u\,du\\
&=\frac{45}{4} \sum_{n=0}^\infty\frac{1}{(n+1)^4}\\
&=\frac{45}{4}\zeta(4)\\
\end{align}
therefore,
\begin{align}A&=\frac{K}{\pi}\\
&=\frac{45}{4\pi}\zeta(4)\end{align}
\begin{align}
B&=\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin(2x)}{2}\right)\,dx
\end{align}
Perform the change of variable $y=2x$,
\begin{align}
B&=\frac{1}{2}\int_0^{\pi}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)\,dx+\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
\end{align}
In the latter integral perform the change of variable $y=\pi-x$,
\begin{align}
B&=\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)\,dx\\
&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin x\right)\,dx-2\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin x)\,dx+\frac{\pi}{2}\ln^2 2\\
&=J+\frac{3\pi}{2}\ln^2 2
\end{align}
Therefore,
\begin{align}\frac{45}{4}\zeta(4)+J+\frac{3\pi}{2}\ln^2 2=4J
\end{align}
Thus,
\begin{align}J&=\frac{15}{4\pi}\zeta(4)+\frac{1}{2}\pi\ln^2 2
\end{align}
If you know that,
\begin{align}\zeta(4)=\frac{\pi^4}{90}\end{align}
therefore,
\begin{align}\boxed{J=\frac{\pi^3}{24}+\frac{1}{2}\pi\ln^2 2}
\end{align}
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