Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3),  $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ?  $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{\sin x} dx = 7~\zeta(3) ,\text { equivalently }\  \int_{0}^{1}\frac{x - x^{2} }{ \sin(\pi x)}dx = 7\frac{\zeta (3)}{\pi^{3}},$$ has been solved here.)

Letting $x\mapsto \frac{\pi}{2}-x$ changes
$$
\begin{aligned}
I=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)^{2}\left(\frac{\pi}{2}+x\right)^{2}}{\cos ^{2} x} d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi^{2}}{4}-x^{2}\right)^{2} d(\tan x)
\end{aligned}
$$
Integration by parts gives
$$
\begin{aligned}
I &=4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) \tan x d x=-8 \int_{0}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) d(\ln (\cos x))
\end{aligned}
$$
Integration by parts again yields
$$
I=2 \pi^{2} \int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x-24 \int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x
$$
Using my post 1 and post 2, we can conclude that
$$
I=-\pi^{3} \ln 2-24\left(-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta (3)\right)=\frac{\pi}{6} \zeta(3)
$$
