Proving $\frac{π^2}8=\sum\limits_{n=1}^\infty\frac1{(2n - 1)^2}$ using Fourier series

Question

I found the Fourier series of $x^2$, which is $$x^2 = \frac{π^2}{3} + 4 \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} \cos(nx).$$ So now, how can I prove that $$\frac{π^2}{8} = \sum_{n = 1}^\infty \frac{1}{(2n - 1)^2}? $$

Answer 1

Hint: first set $x=\pi$ in the Fourier series and show that $\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{π^2}{6}$.