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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.

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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 $$f(x)=\sum_{j=0}^\infty b_{2j+1}\sin\bigl((2j+1)x\bigr)={8\over\pi}\sum_{j=0}^\infty{\sin\bigl((2j+1)x\bigr)\over(2j+1)^3} \qquad(0\leq x\leq\pi)\ .$$

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