Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx =  \frac{7 \zeta(3)}{\pi^3}.
\end{equation}

First, denote the integral below as $I$$$I=\int\limits_0^1dx\space\frac {x(1-x)}{\sin\pi x}$$and through integration by parts on $u=x-x^2$, then we have
$$\begin{align*}I & =-\frac 1{\pi}(x-x^2)\log\cot\left(\frac {\pi x}2\right)\,\Biggr\rvert_0^1+\frac 1{\pi}\int\limits_0^1dx\, (1-2x)\log\cot\left(\frac {\pi x}2\right)\\ & =\frac 1{\pi}\int\limits_0^1dx\,\log\cot\left(\frac {\pi x}2\right)-\frac 2{\pi}\int\limits_0^1dx\, x\log\cot\left(\frac {\pi x}2\right)\\ & =-\frac 8{\pi^3}\int\limits_0^{\pi/2}dx\, x\log\cot x\tag1\end{align*}$$
where equation ($1$) comes from making the substitution $x\mapsto\frac {\pi x}2$. The latter integral can be evaluated by splitting up the natural log into two separate integrals and using the Fourier series for $\log\sin x$ and $\log\cos x$, which I have included below

$$\begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos2kx}{k}-\log 2\tag2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\tag3\end{align*}$$

Expanding ($1$) gives
$$I=-\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\cos x}_{I_1}+\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\sin x}_{I_2}\tag4$$
Call the first and second integrals $I_1$ and $I_2$ respectively. Using ($2$) and ($3$) gives the following two identities
$$\begin{align*}I_1 & =\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {(-1)^{k-1}\cos 2kx}k-x\log 2\right)\\ & =\sum\limits_{k\geq1}\frac {(-1)^{k-1}}k\left[\frac {\pi}{4k^2}\sin\pi k+\frac 1{4k^3}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log2\\ & =\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}\cos\pi k-\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}-\frac {\pi^2}8\log 2\\ & \color{blue}{=-\frac 14\zeta(3)-\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag5\end{align*}$$
As a side note, the infinite sum with $\sin\pi k$ vanishes because $\sin\pi k=0$ for $k\in\mathbb{Z}$. In a similar manner, $I_2$ can be integrated as follows
$$\begin{align*}I_2 & =-\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {\cos 2kx}k+x\log 2\right)\\ & =-\sum\limits_{k\geq1}\frac 1k\left[\frac {\pi}{4k}\sin\pi k+\frac 1{4k^2}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log 2\\ & =-\frac 14\sum\limits_{k\geq1}\frac {\cos\pi k}{k^3}+\frac 14\sum\limits_{k\geq1}\frac 1{k^3}-\frac {\pi^2}8\log 2\\ & \color{red}{=\frac 14\zeta(3)+\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag6\end{align*}$$
Substituting the results for ($5$) and ($6$) into ($4$) leaves us with
$$\begin{align*}I & =-\frac 8{\pi^3}\left[\color{blue}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\color{red}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\right]\\ & =\frac 7{\pi^3}\zeta(3)\end{align*}$$
Multiply by $-1$ to get the integral under question
$$\int\limits_0^1dx\space\frac {x^2-x}{\sin\pi x}\color{brown}{=-\frac 7{\pi^3}\zeta(3)}$$
