Challenging integral: $\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx$ How to tackle
$$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx\ ?$$
This integral popped up in my solution ( see the integral $\mathcal{I_3}\ $ at the end of the solution.)
My attempt:  By Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan^2(x)}{x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=2\int_0^1\frac{\ln(2)+\ln x}{x}\arctan^2(x)dx-2\int_0^1\frac{\ln(1+x^2)}{x}\arctan^2(x)dx$$
The first integral simplifies to known harmonic series using the identity
$$\arctan^2(x)=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}$$
But using this series expansion in the second integral yields very complicated harmonic series. Also integrating by parts, yields the integrand $\frac{\text{Li}_2(-x^2)\arctan(x)}{1+x^2}$ which complicates the problem. Any thought how to approach any of these two integrals?
Thank you.
 A: We have

*

*$\int \frac{\log ^3(1+i x)}{x} \, dx=6 \text{Li}_4(i x+1)+3 \text{Li}_2(i x+1) \log ^2(1+i x)-6 \text{Li}_3(i x+1) \log (1+i x)+\log (-i x) \log ^3(1+i x)$
So

*

*$\Re\left(\int_0^1 \frac{\log ^3(1+i x)}{x} \, dx\right)=\int_0^1 \frac{\frac{1}{8} \log ^3\left(x^2+1\right)-\frac{3}{2} \log \left(x^2+1\right) \tan ^{-1}(x)^2}{x} \, dx\\=-\frac{3}{4} \pi  C \log (2)+\frac{3}{64} \pi  \Im\left(-32 \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\Re\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right) \log (8)-6 \text{Li}_4\left(\frac{1}{2}-\frac{i}{2}\right)\right)-\frac{5}{64} \left(42 \zeta (3) \log (2)+\log ^4(2)\right)+\frac{1249 \pi ^4}{15360}+\frac{21}{128} \pi ^2 \log ^2(2)$
Also one have

*

*$\int_0^1 \frac{\log ^3\left(x^2+1\right)}{x} \, dx=-3 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{8} \zeta (3) \log (8)+\frac{\pi ^4}{30}-\frac{1}{8} \log ^4(2)+\frac{1}{8} \pi ^2 \log ^2(2)$
So

*

*$\int_0^1 \frac{\log \left(x^2+1\right) \tan ^{-1}(x)^2}{x} \, dx=\frac{1}{2} \pi  C \log (2)+\pi  \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{8} \zeta (3) \log (2)-\frac{421 \pi ^4}{11520}+\frac{\log ^4(2)}{24}-\frac{7}{96} \pi ^2 \log ^2(2)$
So

*

*$\int_0^{\frac{\pi }{2}} \frac{x^2 \log (\sin (x))}{\sin (x)} \, dx=-4 \pi  \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{3 \pi ^4}{32}+\frac{1}{8} \pi ^2 \log ^2(2)$
