Consider the $2\pi$-periodic odd function defined on $[0,\pi]$ by $f(x) = x(\pi - x)$ Show that the fourier series of $f$ can be written as $f(x) = \frac{8}{\pi}\sum_{n:odd \geq 1} \frac{sin(nx)}{n^3}$
First I computed $a_0$ to be: $$ \frac{1}{\pi}\int_0^\pi x(\pi - x)dx = \frac{\pi^3}{6}$$
Then since $f$ is odd we get a sine series with coefficients $$b_n = \frac{2}{\pi} \int_0^\pi x(\pi-x)sin(nx)dx = -\dfrac{2\left(\left(2nx-{\pi}n\right)\sin\left(nx\right)+\left(-n^2x^2+{\pi}n^2x+2\right)\cos\left(nx\right)\right)}{{\pi}n^3}_0^\pi$$ $$= -\frac{4}{\pi n^3}$$ where $n$ is odd. I checked the integral with a calculator to make sure it is correct, but it obviously gives us the wrong expression for the fourier series. Any help would be appreciated.