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 $0=0$
 A: 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
$$\int_0^{\pi/2} x^2\log(\sin x)\,dx=  \frac{3\pi}{16}\zeta(3)-\frac{\pi^3}{24}\ln2$$
A: 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 $9.54\times 10^{-10}$.
A: Using this, we have
$$ \log(\sin(x))=-\log 2-\sum_{k=1}^{+\infty}\frac{\cos(2kx)}{k} $$
Therefore,
$$ \int_0^{\pi/2}x^2\log(\sin x)dx=-\frac{\pi^3}{24}\log 2-\sum_{k=1}^{+\infty}\frac{1}{k}\int_0^{\pi/2}x^2\cos(2kx)dx $$
Using integration by parts, we have
$$ \int_0^{\pi/2}x^2\cos(2kx)dx = \frac{(-1)^k\pi}{4k^2} $$
Thus,
$$ \sum_{k=1}^{+\infty}\frac{1}{k}\int_0^{\pi/2}x^2\cos(2kx)dx=\frac{\pi}{4}\sum_{k=1}^{+\infty}\frac{(-1)^k}{k^3}=-\frac{\pi}{4}\eta(3)=-\frac{3\pi}{16}\zeta(3) $$
where $\eta$ is the Dirichlet eta function. Finally,
$$ \int_0^{\pi/2}x^2\log(\sin x)dx=\frac{3\pi}{16}\zeta(3)-\frac{\pi^3}{24}\log 2 $$
