<s>[INST] 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.)
 [/INST] Integrate by parts twice
\begin{align}
\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx 
=& \>2 \int_0^{ \frac\pi2}\frac{x^2(\pi-x)^2}{\sin^2 x} dx \\
=&\>4 \int_0^{\frac\pi2}(\pi^2x-3\pi x^2+2x^3)\cot x\> dx\\ =&\> 4\int_0^{\frac\pi2} (6\pi x-6x^2 -\pi^2)\ln(2\sin x)dx
\end{align}
Then, utilize the known results $\int_0^{\frac\pi2} \ln(2\sin x)dx=0$, $\int_0^\frac{\pi}{2}x\ln(2\sin x)dx= \frac7{16}\zeta(3)$ and $\int_0^{\frac\pi2} x^2\ln(2\sin x)\,dx=  \frac{3\pi}{16}\zeta(3)$ to arrive at
\begin{align}
\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx 
=6\pi\zeta(3),  
\end{align}
</s>