How many ways are there to prove that $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G?$ 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 $$
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G
$$
