How can you approach $int_0^{pi/2} xfrac{ln(cos x)}{sin x}dx$

Here is a new challenging problem:

Show that

$$I=int_0^{pi/2} xfrac{ln(cos x)}{sin x}dx=2ln(2)G-frac{pi}{8}ln^2(2)-frac{5pi^3}{32}+4Imleft{text{Li}_3left(frac{1+i}{2}right)right}$$

My attempt:

With Weierstrass substitution we have

$$I=2int_0^1frac{arctan x}{x}lnleft(frac{1-x^2}{1+x^2}right)dxoverset{xto frac{1-x}{1+x}}{=}4int_0^1frac{frac{pi}{4}-arctan x}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx$$

$$=piunderbrace{int_0^1frac{1}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx}_{I_1}-4underbrace{int_0^1frac{arctan x}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx}_{I_2}$$

By setting $xto frac{1-x}{1+x}$ in the first integral we have

$$I_1=frac12int_0^1frac{1}{x}lnleft(frac{1-x^2}{1+x^2}right)dx$$

$$=frac14int_0^1frac{1}{x}lnleft(frac{1-x}{1+x}right)dx=frac14left[-text{Li}_2(x)+text{Li}_2(-x)right]_0^1=-frac38zeta(2)$$

For the second integral, write $frac{1}{1-x^2}=frac{1}{2(1-x)}+frac{1}{2(1+x)}$

$$I_2=frac12int_0^1frac{arctan x}{1-x}lnleft(frac{2x}{1+x^2}right)dx+frac12int_0^1frac{arctan x}{1+x}lnleft(frac{2x}{1+x^2}right)dx$$

The first integral is very similar to this one

$$int_0^1frac{arctanleft(xright)}{1-x},
lnleft(frac{2x^2}{1+x^2}right),mathrm{d}x =
-frac{pi}{16}ln^{2}left(2right) –
frac{11}{192},pi^{3} +
2Imleft{%
text{Li}_{3}left(frac{1 + mathrm{i}}{2}right)right}$$

So we are left with only $int_0^1frac{arctan xln(1+x^2)}{1+x}dx$ as $int_0^1frac{arctan xln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?

I noticed that if we use $xtofrac{1-x}{1+x}$ in $int_0^1frac{arctan xln(1+x^2)}{1+x}dx$ we will have a nice symmerty but still some annoying integrals appear.

In $I$, I also tried the Fourier series of $ln(cos x)$ but I stopped at $int_0^{pi/2} frac{xcos(2nx)}{sin x}dx$. I would like to see different approaches if possible.

Thank you.