evaluate $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ I am a high school student , I know how to evaluate $\int_0^{\pi/2} x\log(\sin x)\,dx$.
It would be great if someone can help me evaluating $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ and tell me if this integral is elementary or non elementary .
I tried using the "a-x" property but it resulted in $0=0$

Let
$$S=\int_0^{\pi/2} x^2\log(2\sin x)\,dx,\>\>\>\>\>
C=\int_0^{\pi/2} x^2\log(2\cos x)\,dx$$
Since you already knew
$\int_0^{\pi/2} x\log(2\cos x)\,dx=-\frac7{16}\zeta(3)$, apply the variable change $x\to\frac\pi2-x$ to the $S$-integral to get
$$S - C =-\pi\int_0^{\pi/2} x\log(2\cos x)\,dx=\frac{7\pi}{16}\zeta(3)\tag1
$$
Also
\begin{align}
S+C & = \int_0^{\pi/2} x^2\log(2\sin 2x)\,dx
\overset{2x\to x}=\frac18\int_0^{\pi} x^2\log(2\sin x)\,dx\\
&=\frac18S + \frac18\int_{\pi/2}^{\pi} x^2\log(2\sin x)\overset{x\to\frac\pi2+x}{ dx}\\
&= \frac18(S+C) + \frac\pi8 \int_0^{\pi/2} x\log(2\cos x)\,dx
= -\frac\pi{16}\zeta(3)\tag2
\end{align}
Combine (1) and (2) to obtain $S= \frac{3\pi}{16}\zeta(3)$, or
$$\int_0^{\pi/2} x^2\log(\sin x)\,dx=  \frac{3\pi}{16}\zeta(3)-\frac{\pi^3}{24}\ln2$$
