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?
We have that
$$\int_0^{\pi/2}\frac x {\sin x} d x=\int_0^{\pi/4}\frac{2t}{2\sin t \cos t} d (2t)=2\int_0^{\pi/4}\frac{t}{\tan t \cos^2 t}d t=2\int_0^{1}\frac{\arctan s}{s}d s\\
=2\int_0^{1}\sum_{n=0}^{+\infty}\frac{(-1)^n s^{2n}}{2n+1} ds
=2\sum_{n=0}^{+\infty}(-1)^n\int_0^{1}\frac{ s^{2n}}{2n+1} ds
=2\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^2}.$$
Both expressions equal $2G$, where $G$ is the Catalan constant