How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$? Wolfram Alpha shows that $$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$
I tried to use the Fourier series 
$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$
I am not sure how to continue from this point. I need some help.
 A: HINT
By using the basic trigonometric identity
$$1-\cos(x)=2\sin^2\left(\frac x2\right)$$
yor given integral becomes
$$\begin{align}
\int_{0}^{2\pi}x^2\ln (1-\cos x)~dx &= \int_{0}^{2\pi}x^2 \ln\left(2\sin^2\left(\frac x2\right)\right)~dx\\
&=\int_{0}^{2\pi}x^2\ln(2)~dx+2\int_{0}^{2\pi}x^2 \ln\left(\sin\left(\frac x2\right)\right)~dx\\
&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx
\end{align}$$
where within the second integral the substitution $x=\frac x2$ was used. Now use the Fourier series expansion 
$$\ln(\sin x)=-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$
to further get
$$\begin{align}
\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\left[-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right]~dx\\
&=\frac{8\pi^3}{3}\ln(2)-16\int_0^{\pi}x^2\ln(2)~dx-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx
\end{align}$$
The second integral can be evaluated by applying integration by parts. Can you finish it from hereon?
A: Thanks mrtaurho, I am able to finish it.
$$\begin{align}
\int_{0}^{2\pi}\pi^2\ln(1-\cos x)dx
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\left[-x^2\frac{\sin(2nx)}{2n}+\frac{2x\cos(2nx)}{4n^2}-\frac{2\sin(2nx)}{8n^3}\right]_{0}^{\pi}\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}{n}\frac{2\pi}{4n^2}\\
&= -\frac{8\pi^3}3\ln(2)-8\pi\sum_{n=1}^{\infty}\frac{1}{n^3}\\
&=-\frac{8}{3}\pi(\pi^2\ln2+3\zeta(3))
\end{align}$$
