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}

$\displaystyle f(a):=\int\limits_0^1 x e^{ax}dx = \frac{1+e^a(a-1)}{a^2}$
$\displaystyle g(a):=\int\limits_0^1 x^2 e^{ax}dx = \frac{-2+e^a(a^2-2a+2)}{a^3}$
$\displaystyle \int\limits_0^1 \frac{x^2-x}{\sin(\pi x)}dx = i2\int\limits_0^1\frac{x^2-x}{e^{i\pi x}-e^{-i\pi x}}dx = i2\sum\limits_{k=0}^\infty \int\limits_0^1 (x^2-x)e^{-i\pi x(2k+1)}dx $
$\displaystyle = i2\sum\limits_{k=0}^\infty (g(-i\pi(2k+1))-f(-i\pi(2k+1)))$
$\displaystyle = 2\sum\limits_{k=0}^\infty i\frac{-2+i\pi(2k+1) + e^{-i\pi(2k+1)}(2+i\pi(2k+1))}{(-i\pi(2k+1))^3} \enspace$ with $\enspace e^{-i\pi(2k+1)}=-1$
$\displaystyle = -8\sum\limits_{k=0}^\infty\frac{1}{(\pi(2k+1))^3}=-\frac{8}{\pi^3}(1-\frac{1}{2^3})\zeta(3)=-\frac{7\zeta(3)}{\pi^3}$
