# Fourier sine expansion for $x(\pi - x)$

I'm trying to find the Fourier sine expansion for the function $$f(x)$$ = $$x(\pi - x)$$ for the interval $$0 \leq x \leq \pi$$.

I think I am supposed to find $$\sum_{k=1}^{n} b_k \sin{(kx)}$$ where $$b_k = \frac{2}{\pi}$$ $$\int_{0}^{\pi} x(\pi - x) \sin{(kx)} dx$$. However this is turning out really complex and nothing like the answer in my textbook, which is $$f(x) = \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{\sin{(2k + 1)x}}{(2k + 1)^3}$$.

Any help is appreciated.

Your formula for $$b_k$$ is correct in this example. Compute the integral correctly, and you will see that you obtain $$b_k=0$$ when $$k$$ is even, while $$b_{2j+1}={8\over\pi(2j+1)^3}\qquad(j\geq 0)\ .$$ This leads to $$f(x)=\sum_{j=0}^\infty b_{2j+1}\sin\bigl((2j+1)x\bigr)={8\over\pi}\sum_{j=0}^\infty{\sin\bigl((2j+1)x\bigr)\over(2j+1)^3} \qquad(0\leq x\leq\pi)\ .$$