Can you help me please?
PROBLEM: Expand the function $y = x (π - x)$ in the sine Fourier series on the segment $[0, π]$. Using this result, find the sum of the series
$$\sum _{n=0}^{\infty \:}\:\frac{\left(-1\right)^n}{\left(2n-1\right)^3}$$
MY ATTEMPT:
\begin{align} L&=\frac{b-a}{2}\\ a_0&=\frac{2}{\pi }\int _0^{\pi }x\left(\pi -x\right)\,dx=\frac{\pi^{2} }{3}\\ a_n&=\frac{2}{\pi }\int _0^{\pi }x\left(\pi -x\right)\cos\left(2nx\right)\,dx=\frac{\cos(n\pi)(\sin(n\pi)-\pi n\cos(n\pi))}{\pi n^3}\\ b_n&=\frac{2}{\pi }\int _0^{\pi }x\left(\pi -x\right)\sin\left(2nx\right)\,dx= \frac{\sin(n\pi)(\sin(n\pi)-\pi n\cos(n\pi))}{\pi n^3} \end{align}
What do I do next, or why am I wrong? Thanks