In my post, I had found the values of the couple of integrals, $$ \int_{0}^{\frac{\pi}{4}} x \tan x d x=-\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(1)} $$ and $$ \int_{0}^{\frac{\pi}{4}} x \cot x d x=\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(2)} $$ I just wonder what happens when I combine these 2 results and consequently find a wonderful result $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G. $$
Adding (1) and (2) yields $$ G=\int_{0}^{\frac{\pi}{4}} x \tan x d x+ \int_{0}^{\frac{\pi}{4}} x \cot x d x$$
$$ \begin{aligned} \because x \tan x+x \cot x &=x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) \\ &=\frac{x}{\cos x \sin x} \\ &=\frac{2 x}{\sin (2 x)} \end{aligned} $$
$$ \therefore \int_{0}^{\frac{\pi}{4}} x \tan x d x+\int_{0}^{\frac{\pi}{4}} x \cot x d x=\int_{0}^{\frac{\pi}{4}} \frac{2 x}{\sin (2 x)} d x $$Letting $x\mapsto \frac{x}{2}$ yields $$ G=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x $$ Now we can conclude that $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G. $$
I am curious whether there are more elegant proofs.
Your comments and alternate solutions are warmly welcome.
Inspired by Mr Svyatoslav,
$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x &=4 \int_{0}^{\frac{\pi}{2}} x d\left(\ln \left(\tan \frac{x}{2}\right)\right) \\ &=4\left[x \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-4 \int_{0}^{\frac{\pi}{2}} \ln \left(\tan \frac{x}{2}\right) d x \\\\ &=-2 \int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x \end{aligned} $$
By my post $$ \int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x=-G,$$
we can conclude that $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G $$