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.)

A more general solution:
I’ll state the following propositions that can be trivially proven using induction, sums of geometric series, second derivatives and the use of $\tan^2 x=\sec^2 x-1$.

For $n\in\mathbb{N}$
$$\sec^{2n}(x)=\frac{(-1)^n \, 2^{2n}}{(2n-1)!} \sum_{k=1}^{\infty} (-1)^k e^{2ixk} \prod_{r=0}^{2n-2} (k-n+r+1)$$
$$\sec^{2n+1}(x)=\frac{(-1)^{n-1} \, 2^{2n+1}}{(2n)!} \sum_{k=1}^{\infty} (-1)^k e^{ix(2k-1)} \prod_{r=0}^{2n-1} (k-n+r)$$
Letting $x\mapsto \frac{\pi}{2}-x$, we recover representations of powers of $\csc x$.

To illustrate an example:
$$\csc^2 (x)= -4\sum_{k=1}^{\infty} (-1)^k k e^{2i\left(\frac{\pi}{2}-x\right)k}$$
Let’s use this to integrate
$$\begin{align} \int_{0}^{\pi} x^2 (\pi-x)^2 \csc^2 (x)\,dx &= -4\sum_{k=1}^{\infty} (-1)^k k \int_{0}^{\pi} x^2(\pi^2-x)^2 e^{2i\left(\frac{\pi}{2}-x\right)k}\,dx\\&=-4\sum_{k=1}^{\infty} (-1)^k k \cdot \left(-\frac{3\pi \, (-1)^k}{2k^4}\right)\\&=6\pi \sum_{k=1}^{\infty}\frac{1}{k^3}\\&=6\pi \zeta(3)\end{align}$$
