How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx\overset{x\to \frac{1-x}{1+x}}{=}4\int_0^1\frac{\frac{\pi}{4}-\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=\pi\underbrace{\int_0^1\frac{1}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_1}-4\underbrace{\int_0^1\frac{\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_2}$$
By setting $x\to \frac{1-x}{1+x}$ in the first integral we have
$$I_1=\frac12\int_0^1\frac{1}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$=\frac14\int_0^1\frac{1}{x}\ln\left(\frac{1-x}{1+x}\right)dx=\frac14\left[-\text{Li}_2(x)+\text{Li}_2(-x)\right]_0^1=-\frac38\zeta(2)$$
For the second integral, write $\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$
$$I_2=\frac12\int_0^1\frac{\arctan x}{1-x}\ln\left(\frac{2x}{1+x^2}\right)dx+\frac12\int_0^1\frac{\arctan x}{1+x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
The first integral is very similar to this one
$$\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$
So we are left with only $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ as $\int_0^1\frac{\arctan x\ln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?
I noticed that if we use $x\to\frac{1-x}{1+x}$ in $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ 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 $\int_0^{\pi/2} \frac{x\cos(2nx)}{\sin x}dx$. I would like to see different approaches if possible.
Thank you.

Many ways to go are possible!
A simple way would be to exploit the known result,
$$\int_0^1 \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)=\frac{\pi^3}{16},\tag 1$$
since with the Weierstrass subs the main integral reduces to
$$\mathcal{I}=2\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1-x^2}{1+x^2}\right)\textrm{d}x$$
$$=-2 \int_0^1 \frac{ \arctan(x)}{x}\log \left(\frac{1+x^2}{(1-x)^2}\right) \textrm{d}x-2 \int_0^1 \frac{\arctan(x)\log (1-x)}{x} \textrm{d}x$$
$$+2 \int_0^1 \frac{\arctan(x)\log (1+x) }{x} \textrm{d}x$$
$$=2\log(2)G-\frac{\pi}{8}\log^2(2)-\frac{5}{32}\pi^3+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\},$$
where the last two integrals are calculated by Ali Shather in this answer https://math.stackexchange.com/q/3261446.
End of story
Credit for this approach goes to Cornel.
A first note: Interestingly, different ways make the problem very difficult. It would be nice to have in place more ways to go.
A second note: The generalization of the key integral in $(1)$ may be found in the book, (Almost) Impossible Integrals, Sums, and Series, page $17$,
$$ \int_0^x \frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt)\log (1-t)}{t}\textrm{d}t$$
$$=2\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3}, \ |x|\le1.$$
