Find: $\int_0^\pi x^2\ln(\sin(x))dx$ I've been working on a few log-sine integrals. So far I have found
$$\int_0^\pi \ln(\sin(x))dx=-\pi\ln(2)$$
$$\int_0^\pi x\ln(\sin(x))dx=-\frac{\pi^2\ln(2)}{2}$$
...but I am struggling with the integral
$$\int_0^\pi x^2\ln(\sin(x))dx$$
and I can't figure it out... however, I do know that the answer will contain $\zeta(3)$. Any hints?
 A: Hint: we know 
$$\int_a^b f(x)dx= \int_a^b f(a+b-x)dx$$
With $a=0,~~~b=\pi$ set 
$$I=\int_0^\pi \ln(\sin(x))dx~~~~and ~~~~~
J=\int_0^\pi x\ln(\sin(x))dx$$$$A=\int_0^\pi x^3\ln(\sin(x))dx~~~~~and~~~~~B =\int_0^\pi x^2\ln(\sin(x))dx $$
From above formula  we have 
\begin{split}\int_0^\pi x^3\ln(\sin(x))dx &=&-\int_0^\pi (x-\pi)^3\ln(\sin(\pi-x))dx \\
&=&-\int_0^\pi x^3\ln(\sin(x))dx +3\pi \int_0^\pi x^2\ln(\sin(x))dx \\&-&3\pi^2\int_0^\pi x\ln(\sin(x))dx+\pi^3 \int_0^\pi \ln(\sin(x))dx\end{split}
that is $$2A -3\pi B = -3\pi^2J+\pi^3 I$$
Do this again with 
$$\int_0^\pi x^4 \ln(\sin(x))dx =\int_0^\pi (x-\pi)^4\ln(\sin(\pi-x))dx$$ 
Which is equivalent after doing as above to 
$$ -4\pi A+6\pi^2B -4\pi^3J+\pi^4 I=0$$
then you will get A and B by solving 
$$2A -3\pi B = -3\pi^2J+\pi^3 I$$
$$ -4\pi A+6\pi^2B =4\pi^3J-\pi^4 I$$
A: Rewrite the integral as:
$$\begin{aligned}\int_0^\pi  {{x^2}\ln (\sin x)dx}  &= \int_{ - \pi /2}^{\pi /2} {{{(x + \frac{\pi }{2})}^2}\ln (\cos x)dx}  \\ &= 2\int_0^{\pi /2} {{x^2}\ln (\cos x)dx}  + \frac{{{\pi ^2}}}{2}\underbrace{\int_0^{\pi /2} {\ln (\cos x)dx}}_{-\pi \ln 2 / 2} \end{aligned}$$
To evaluate the first integral, we use the Fourier expansion of $\ln(\cos x)$:
$$\ln (\cos x) =  - \ln 2 - \sum\limits_{k = 1}^\infty  {\frac{{{{( - 1)}^k}\cos 2kx}}{k}} $$
Plug it into the integral (it is legitimate to interchange order of summation and integration, because the resulting series is absolutely convergent), we obtain
$$\begin{aligned}
\int_0^{\pi /2} {{x^2}\ln (\cos x)dx} &= - \ln 2\int_0^{\pi /2} {{x^2}dx}  - \sum\limits_{k = 1}^\infty  {\frac{{{{( - 1)}^k}}}{k}\int_0^{\pi /2} {{x^2}\cos 2kxdx} }  \\& =  - \frac{{{\pi ^3}\ln 2}}{{24}} - \sum\limits_{k = 1}^\infty  {\frac{{{{( - 1)}^k}}}{k}\frac{{\pi {{( - 1)}^k}}}{{4{k^2}}}}  \\ &=  - \frac{{{\pi ^3}\ln 2}}{{24}} - \frac{\pi }{4}\zeta (3) \end{aligned}$$
The final result is, $$\int_0^\pi  {{x^2}\ln (\sin x)dx}  = -\frac{\pi^3}{3}\ln 2 -\frac{\pi}{2}\zeta(3)$$
