$\int_0^{\pi/4} (\ln\sin x)^2 \, dx$ Help me to integrate this, please. I tried to do this one for three days, but I have no idea.
$$
\int_0^{\pi/4} (\ln\sin x)^2\, dx
$$
 A: $$\mathfrak{I}=\int_{0}^{\pi/4}\log^2(\sin x)\,dx = \int_{0}^{1/\sqrt{2}}\frac{\log^2(x)}{\sqrt{1-x^2}}\,dx=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\int_{0}^{1/\sqrt{2}}x^{2n}\log^2(x)\,dx $$
just depends on the following hypergeometric series:
$$ S_3 = \sum_{n\geq 0}\frac{\binom{2n}{n}}{8^n(2n+1)^3}={}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right) $$
$$ S_2 = \sum_{n\geq 0}\frac{\binom{2n}{n}}{8^n(2n+1)^2}=\frac{4G+\pi\log 2}{4\sqrt{2}} $$
$$ S_1 = \sum_{n\geq 0}\frac{\binom{2n}{n}}{8^n(2n+1)}=\frac{\pi}{2\sqrt{2}} $$
$S_3$ has been evaluated at page 39 here through Fourier-Legendre series expansions. It depends on $\pi,G,\log 2$ and $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$. It was previously computed by Random Variable and others in this historical thread.
A: As $\sin x\approx x$ the best I could do was to approximate the integral to
$$\int_0^{\pi/4} \log^2 x \,dx=\lim_{\varepsilon\to 0}\left[2 x+x \log ^2 x-2 x \log x\right]_{\varepsilon}^{\pi/4}=$$
$$=\frac{\pi}{2}+\frac{\pi}{4}\log^2\frac{\pi}{4}-\frac{\pi}{2}\log\frac{\pi}{4}-\lim_{\varepsilon\to 0} (2 \varepsilon+\varepsilon \log ^2 \varepsilon-2 \varepsilon \log \varepsilon)\approx 1.996 $$
