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}

An elementary evaluation:
$$\int_0^1 \frac{x-x^2}{\sin \pi x} dx \overset{ibp}
=\int_0^1 \frac{2x-1}\pi\ln \tan \frac{\pi x}2 dx
\overset{t=\tan^2\frac{\pi x}2}=
\frac2{\pi^3}\int_0^\infty \frac{\ln t\tan^{-1}\sqrt t}{\sqrt t(1+t)}dt
$$
Let $J(a)=\int_0^\infty \frac{\ln t\tan^{-1}a\sqrt t}{\sqrt t(1+t)}dt$
$$J’(a)= \int_0^\infty \frac{\ln t \>dt}{(1+t)(1+a^2t)}
\overset{y=\frac1{a^2t}}=- \ln a\int_0^\infty \frac{ dy}{(1+y)(1+a^2y)}= \frac{2\ln^2a}{1-a^2}
$$
Then
$$I = \frac2{\pi^3}J(1)= \frac2{\pi^3}\int_0^1 J’(a)da
=\frac4{\pi^3}\int_0^1 \frac{\ln^2a}{1-a^2} da
= \frac4{\pi^3}\frac{7\zeta(3)}4
=\frac{7 \zeta(3)}{\pi^3}
$$
