Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$

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

$$\int\limits_0^1dx\space\frac {x^2-x}{\sin\pi x}\color{brown}{=-\frac 7{\pi^3}\zeta(3)}$$

$\displaystyle f(a):=\int\limits_0^1 x e^{ax}dx = \frac{1+e^a(a-1)}{a^2}$

$\displaystyle g(a):=\int\limits_0^1 x^2 e^{ax}dx = \frac{-2+e^a(a^2-2a+2)}{a^3}$

$\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 $\enspace e^{-i\pi(2k+1)}=-1$

$\displaystyle = -8\sum\limits_{k=0}^\infty\frac{1}{(\pi(2k+1))^3}=-\frac{8}{\pi^3}(1-\frac{1}{2^3})\zeta(3)=-\frac{7\zeta(3)}{\pi^3}$

Probably not an answer.

For the antiderivative $$I=2 \pi^3\int \frac{x^2-x}{\sin (\pi x)}\, dx$$ a CAS give the ugly $$I=-i \pi (2 x-1) \left(4 \text{Li}_2\left(e^{i \pi x}\right)-\text{Li}_2\left(e^{2 i \pi x}\right)\right)+8 \text{Li}_3\left(e^{i \pi x}\right)-\text{Li}_3\left(e^{2 i \pi x}\right)-$$ $$4 \pi ^2 (x-1) x \tanh ^{-1}\left(e^{i \pi x}\right)$$ $$\lim_{x\to 1} \, I=-7 \zeta (3)+i\frac{ \pi ^3}{2} \qquad \text{and} \qquad \lim_{x\to 0} \, I=7 \zeta (3)+i\frac{ \pi ^3}{2}$$

What is interesting is that a rather good approximation could be obtained using a $[2,2]$ Padé approximant built at $x=\frac 12$ making $$\frac{x^2-x}{\sin (\pi x)}=\frac{-\frac 14+ a(x-\frac 12)^2 }{1+ b(x-\frac 12)^2 }$$ where $$a=-\frac{384-48 \pi ^2+\pi ^4}{48 \left(\pi ^2-8\right)} \qquad \text{and} \qquad b=-\frac{5 \pi ^4-48 \pi ^2}{12 \left(\pi ^2-8\right)}$$ making the definite integral easy to solve (leading to a value of $\approx -0.271415$ while the exact value is $\approx -0.271377$).

Still more amazing (at least to me), the approximation $$\sin(y) \simeq \frac{16 (\pi -y) y}{5 \pi ^2-4 (\pi -y) y}\qquad (0\leq y\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician would lead to $-\frac{13}{48} \approx -0.270833$ .