In this recent paper on log tangent integrals, Theorem 1 (2.2, in the published version) expresses that, if $n$ is a positive integer, \begin{equation} \int_0^{\tfrac{\pi}{2}}E_{2n-1}\left( \frac{2}{\pi}x \right)\log(\tan x)\,dx= \frac{(-1)^{n-1}(2n-1)!}{\pi^{2n-1}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} where $E_n(x)$ are the Euler polynomials. This expression can be written as \begin{equation} \int_0^{1}E_{2n-1}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx= \frac{(-1)^{n-1}2(2n-1)!}{\pi^{2n}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} Defining the antiderivatives \begin{equation} F_{2n-1}(x)=\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} one may notice that $F_{2n-1}(0)=F_{2n-1}(1)=0$, as $E_{2n-1}( 1-x )=-E_{2n-1}( x )$. Then, integrating by parts, it comes \begin{equation} \int_0^1\frac{ F_{2n-1}(x)}{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} As $x=0,1$ are two roots of $F_{2n-1}(x)$, we conclude that the polynomials \begin{equation} f_{2n-1}(x)=\frac{1}{x(1-x)}\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} verify \begin{equation} \int_0^1 f_{2n-1}(x)\frac{x(1-x) }{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} More generally, for symmetry reasons, any polynomial \begin{equation} g_{2n-1}(x)=f_{2n-1}(x)+P(2x-1) \end{equation} where $P(x)$ is an arbitrary odd polynomial, gives the same result. This result gives an explicit representation of the polynomials derived in my previous answer.
Edit: Using the derivative property for the Euler polynomials, $E_{2n-1}(x)=(2n)^{-1}dE_{2n}(x)/dx$, one can express \begin{align} f_{2n-1}(x)&=\frac{1}{2n}\frac{1}{x(1-x)}\left[E_{2n}(x)-E_{2n}(0)\right]\\ &=\frac{1}{x(1-x)}\left[\frac{1}{2n}E_{2n}(x)+\frac{1}{n(2n+1)}\left( 2^{2n+1}-1 \right)B_{2n+1}\right] \end{align} where $B_{2n+1}$ is a Bernoulli number.