This is inspired by the same beautiful integral expression for $\zeta(3)$ as this question, but goes in a slightly different direction. Writing the original integral in the form $$\int_0^1\frac{x(1-x)}{\sin\pi x}dx=7\frac{\zeta(3)}{\pi^3} ,$$ it turns out that for $n\in\mathbb N$ there is a unique monic polynomial $p_n$ of degree $n-1$ such that $$\int_0^1x^np_n(x)\frac{1-x}{\sin\pi x}dx=c_{2n+1}\frac{\zeta(2n+1)}{\pi^{2n+1} } $$ with rational $c_k=4(k-1)! \dfrac{2^k-1}{2^k}= (4-2^{2-k})(k-1)! $.

This follows for $\zeta(2n+1)$ from solving the linear system given by the blue lines numbered $n+1,...,2n$ in the other question. As Zurab Silagadze answered it by giving an explicit formula of the coefficients in the blue lines, the $p_n$'s can be calculated. I don't know however if it is possible to give a formula in *closed* form (meaning here that it should not contain a matrix inversion), but see below.

The first polynomials are $$\begin{align} p_1(x)&=1 \\ p_2(x)&=3-x \\ p_3(x)&=25-20x+x^2 \\ p_4(x)&=455-707x+287x^2-x^3 \\ p_5(x)&=14301-34734x+29046x^2-8304x^3+x^4 \\ p_6(x)&=683067-2289309x+2949276x^2-1721434x^3+382547x^4-x^5 \\ \end{align}$$

The constant terms are supposedly the sequence A272482, thus, correcting the oeis typo $1/(2n)!$, $$[x^0]p_n(x)= {(2n)!}[x^{2n}y^n]\frac{\cos\frac{x(1-y)}{2}} {\cos\frac{x(1+y)}{2}} = \frac 1{4^n} {2n\choose n}\sum_{i=0}^n{n\choose i}E_i,$$ where $E_i$ are the Euler numbers. This seems to suggest something similar for the other coefficients, and thus possibly a closed form.

Are the $p_n$ known? How to find their closed form or generating function?

More generally now, define $$J(m,n,k)=J(n,m,k):= \int_0^1\frac{x^m(1-x)^n}{\sin^k\pi x}dx.$$For this to converge, we need $m,n\geqslant k$.

Experimentally, the situation for $k=2$ is quite similar to the $k=1$ case in that $J(m,n,k)$ is a rational combination of values $\dfrac{\zeta(i)}{\pi^{i+1}}$ with $i$ running over all odd numbers between $\min(m,n)$ and $m+n-1$, e.g. $$J(7,4,2)=\dfrac{105}2\left(-\dfrac{\zeta(5)}{\pi^6}+51\dfrac{\zeta(7)}{\pi^8}-405\dfrac{\zeta(9)}{\pi^{10}}\right).$$

**This leads to new possibilities of representing odd zeta values as integrals, this time with $\sin^2\pi x$ in the denominator.** Writing as a shortcut $h_m:=J(m,m,2)$, we can for example express $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ as a rational combination of $h_2,\dots,h_n$, i.e. as an integral $$\dfrac{\zeta(2n-1)}{\pi^{2n}}=\int_0^1q_n(x-x^2)\frac{x^2(1-x)^2}{\sin^2\pi x}dx,$$ where $q_n$ is a unique polynomial of degree $n-2$. The first of them are:
$$\begin{align} \zeta(3)&=\frac{\pi^4}{6}h_2 \\
\zeta(5)&=\frac{\pi^6}{90}(h_2 +2h_3),\qquad \text{ i. e. } q_2(z)=\frac{1}{90}(1+2z) \quad \text{ etc. }\\
\zeta(7)&=\frac{\pi^8}{1890}(2h_2 +4h_3+3h_4)\\
\zeta(9)&=\frac{\pi^{10}}{28350}(3h_2 +6h_3+5h_4+2h_5)\\
\zeta(11)&=\frac{\pi^{12}}{935550}(10h_2 +20h_3+17h_4+8h_5+2h_6)\\
\zeta(13)&=\frac{\pi^{14}}{638512875}(\color{blue}{691}h_2 +1382h_3+1180h_4+574h_5+175h_6+30h_7)\\ \end{align}$$

Experimentally, in $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ the last coefficient (i.e. the one of $h_n$ and the leading term of $q_n$) is $\dfrac{2^{2n-2}}{(2n)!}$ and the one preceding it is $\dfrac{n(n-2)}6\dfrac{2^{2n-2}}{(2n)!}$, while for the first coefficient (equally, the constant term of $q_n$), the occurrence of $\color{blue}{691}$ in the expression for $\zeta(13)$ suggests that it involves the Bernoulli number $B_{2n-2}$.

Any ideas about these polynomials?

Finally, for $k\geqslant 3$ there does not seem to exist any closed form, at least not in terms of zeta values.

What about $J({3,3,3})= \int\limits_0^1\dfrac{x^3(1-x)^3}{\sin^3\pi x}dx$?

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