Conjecture about equality of integrals with specific 2nd and 3rd order polynomial multiplying a function of $\sin^2(\pi t)$ Is it true that, for a general class of functions $f,$
$$2\int_0^1(y^3 -\frac{1}{4})f(\sin^2{(\pi \, y)})\,dy \, \overset{?}{=} \, \int_0^1(1-y)(1-3y)f(\sin^2{(\pi \, y)})\,dy$$
The above conjecture has been proved true for
$$f(\sin^2{(\pi \, y)}) = \log{\big(1+\frac{4x}{(1-x)^2}\sin^2{(\pi \, y)} \big)} $$ and the closed-form evaluation is $\frac{3}{\pi^2} \text{Li}_3(x),$ where the function mentioned is the trilogarithm.  The surprising thing was that I found the same answer for a cubic and quadratic polynomial.  I thus began testing other continuous functions to gather evidence for the conjecture.  These functions include polylogarithms, Bessel functions of different orders, and exponentials, and so far all functions entered have numerically integrated to be the same on both sides of the proposed equation.
The proposed equation looks like the next in line from the known equation,
$$2\int_0^1(y)f(\sin^2{(\pi \, y)})\,dy \, =\, \int_0^1(1)f(\sin^2{(\pi \, y)})\,dy.$$
Therefore, a logical generalization is, can polynomials $p_n(y)$ and $q_n(y),$ each of degree $n,$ be found such that
$$\int_0^1p_{n+1}(y))f(\sin^2{(\pi \, y)})\,dy \, = \, \int_0^1q_n(y)f(\sin^2{(\pi \, y)})\,dy$$ for sufficiently 'nice' functions?  It would not surprise me that this problem has a known solution, and I would like a reference, if possible.
 A: This identity can be shown as follows
\begin{align}
 I_{\text{left}}&=\int_0^1(y^3 -\frac{1}{4})f(\sin^2{(\pi \, y)})\,dy \\
 &\overset{z=1-y}{=} \int_0^1 \left(-z^3 + 3z^2 - 3z + \frac34\right)f(\sin^2{(\pi \,z)})\,dz\\
 &=-\int_0^1 \left(z^3 -  \frac14\right)f(\sin^2{(\pi \,z)})\,dz+\int_0^1 \left( 3z^2 - 3z + \frac12\right)f(\sin^2{(\pi \,z)})\,dz
\end{align}
and thus,
\begin{align}
 2I_{\text{left}}&=\int_0^1 \left( 3z^2 - 4z + 1\right)f(\sin^2{(\pi \,z)})\,dz+\int_0^1 \left( z-\frac12\right)f(\sin^2{(\pi \,z)})\,dz\\
 &=\int_0^1 (1-z)(1-3z)f(\sin^2{(\pi \,z)})\,dz+\int_0^1 \left( z-\frac12\right)f(\sin^2{(\pi \,z)})\,dz\\
 &=I_{\text{right}}+\int_0^1 \left( z-\frac12\right)f(\sin^2{(\pi \,z)})\,dz
\end{align}
Changing $z=1-u$ in the latter integral shows that it vanishes.
More generally, if
\begin{equation}
 R_{2n}(y)=P_{2n+1}(y)+P_{2n+1}(1-y)+a\left(y-\frac12\right)
\end{equation}
where $a$ is a constant and the degrees of the polynomials ar indicated by the indices, then
\begin{equation}
 2\int_0^1P_{2n+1}(y)f(\sin{(\pi \, y)})\,dy=\int_0^1R_{2n}(y)f(\sin{(\pi \, y)})\,dy  
\end{equation}
