Find Fourier half range sine Series for $f(x)=x(\pi-x)$ on $(0,\pi)$.

Question

I tried finding the Fourier coefficient $ b_n$ but it came zero.

My attempt:

$ b_n=\frac{4}{\pi}\int_0^\pi x(\pi-x)\sin(2nx)dx\\ =\frac{4}{\pi}\left\{\pi\int_0^\pi x\sin(2nx)dx-\int_0^\pi x^2\sin(2nx)dx\right\}\\ =0$

Correct answer is:

$\frac{8}{\pi}\left(\frac{\sin x}{1^3}+\frac{\sin3x}{3^3}+\frac{\sin5x}{5^3}+...\right)$

Answer 1

You should integrate $$\frac{2}{\pi} \int_0^{\pi } x (\pi -x) \sin (n x) \, dx=\frac{4-2 \pi n \sin (\pi n)-4 \cos (\pi n)}{\pi n^3}$$ to get coefficients $$\frac{8}{\pi },0,\frac{8}{27 \pi },0,\frac{8}{125 \pi },\ldots$$ and finally $$\frac{8}{\pi}\left(\sin x +\frac{\sin 3x}{3^3}+\frac{\sin 5x}{5^3}\ldots\right)$$

Answer 2

The correct formula for $b_n$ is $$\frac{2}{\pi}\int_0^\pi x(\pi-x)\sin(nx)\,dx$$