Prove $\int_0^{\pi/2}\frac x {\sin x} \, \mathrm d x = 2\sum_{n\mathop = 0}^{\infty} \frac {(-1)^n}{(2n+1)^2}$ I would like to prove that
$$\int_0^{\pi/2}\frac x {\sin x} \, \mathrm d x = 2\sum_{n\mathop = 0}^{\infty} \frac {(-1)^n}{(2n+1)^2}$$
Any hints?
 A: We have that
$$\int_0^{\pi/2}\frac x {\sin x} d x=\int_0^{\pi/4}\frac{2t}{2\sin t \cos t} d (2t)=2\int_0^{\pi/4}\frac{t}{\tan t \cos^2 t}d t=2\int_0^{1}\frac{\arctan s}{s}d s\\
=2\int_0^{1}\sum_{n=0}^{+\infty}\frac{(-1)^n s^{2n}}{2n+1} ds
=2\sum_{n=0}^{+\infty}(-1)^n\int_0^{1}\frac{ s^{2n}}{2n+1} ds
=2\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^2}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Since $\ds{\totald{\ln\pars{\tan\pars{x/2}}}{x} = {1 \over \sin\pars{x}}}$:

\begin{align}
\color{#f00}{\int_{0}^{\pi/2}{x \over \sin\pars{x}}\,\dd x} & =
-\int_{0}^{\pi/2}\ln\pars{\tan\pars{x \over 2}}\,\dd x
\end{align}
With the Weierstrass Tangent Half-Angle Substitution:
\begin{align}
\color{#f00}{\int_{0}^{\pi/2}{x \over \sin\pars{x}}\,\dd x} & =
-2\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x =
-2\sum_{n = 0}^{\infty}\pars{-1}^{n}\
\overbrace{\int_{0}^{1}\ln\pars{x}\,x^{2n}\,\dd x}
^{\ds{-\,{1 \over \pars{2n + 1}^{2}}}} =
\color{#f00}{2\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}}}
\end{align}
A: Both expressions equal $2G$, where $G$ is the Catalan constant
