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 $0=0$

Another possibility is to write
$$\log[\sin(x)]=\sum_{n=1}^\infty (-1)^n \frac{2^{2 n-3}\,\, (E_{2 n-1}(1)-E_{2 n-1}(0))}{n \,(2 n-1)!}\left(x-\frac{\pi }{2}\right)^{2 n}$$ where appear  the Euler polynomial and use
$$\int_0^{\frac \pi 2}x^2\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\frac{\left(\frac{\pi }2\right)^{2 n+3}}{(n+1) (2 n+1) (2 n+3)}$$ This gives
$$I_2=\frac 1{16}\sum_{n=1}^\infty (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}$$
which, after a series of tedious manipulations, leads to the result already given by @Tuvasbien.
Numerically, this summation does not converge quite fast. Considering the partial (from $n=1$ to $n=p$)
$$\left(
\begin{array}{cc}
 p &  \Sigma_p \\
 1 & -0.159385 \\
 2 & -0.178112 \\
 3 & -0.183246 \\
 4 & -0.185204 \\
 5 & -0.186108 \\
 6 & -0.186581 \\
 7 & -0.186852 \\
 8 & -0.187019 \\
 9 & -0.187127 \\
 10 & -0.187200 \\
\cdots & \cdots \\
\infty & -0.187426
\end{array}
\right)$$
You could easily generalized the result for
$$I_p=\int_0^{\frac \pi 2}x^p\log[\sin(x)]\,dx$$ since
$$\int_0^{\frac \pi 2}x^p\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\left(\frac{\pi }{2}\right)^{2 n+p+1}\frac{(2 n)!\,\,p!}{(2 n+p+1)!}$$
Edit
Numerically, if you look at this question of mine, we could have very good approximations writing
$$I_2\sim\frac 1{16}\sum_{n=1}^p (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}-\frac {\pi^3}8\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}$$ and
$$\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}=\frac{1}{3} \left(H_{p+\frac{3}{2}}-H_p\right)-\left(H_{p+1}-H_{p+\frac{1}{2}}\right)$$
Using $p=5$ gives an absolute error equal to $9.54\times 10^{-10}$.
