Here is a new challenging [problem](https://www.facebook.com/photo.php?fbid=3200606550036121&set=a.222846247812181&type=3&theater):

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](https://math.stackexchange.com/questions/3783058/evaluating-int-01-frac-arctan-x-ln-left-frac2x21x2-right1-xdx)

$$\int\_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}\_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$

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](https://math.stackexchange.com/q/1842492). 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.