Integrate $ \int^{\pi}_{-\pi} (\pi^2-x^2)\sin nx \ dx$

Question

Consider the function $f:(-\pi,\pi)\to\mathbb{R}$ be defined as $x \mapsto (\pi+x)(\pi-x)$

Compute the fourier series of $f$

So far, I've worked out $a_o$ by: \begin{equation} a_o = \frac{1}{\pi} \int^{\pi}_{-\pi} (\pi^2-x^2) \ dx \end{equation} \begin{equation} =\frac{4}{3}\pi^2 \end{equation}

And I got stuck working out $a_n$ \begin{equation} \int^{\pi}_{-\pi} (\pi^2-x^2)\cos nx \ dx \end{equation} \begin{equation} = \int^{\pi}_{-\pi} \left\lbrace \left[ (\pi^2-x^2) \frac{1}{n} \sin nx\right]^\pi_{-\pi} - \frac{1}{n} \int^{\pi}_{-\pi} (\pi^2-x^2)\sin nx \ \right \rbrace dx \end{equation}

I'm stuck on integrating the $ \int^{\pi}_{-\pi} (\pi^2-x^2)\sin nx \ dx$ par

Answer 1

a quick answer, hope this helps:

using integration by parts you obtain: \begin{align} \int x^2\sin(nx)dx&=\frac{2x\sin nx}{n^2}-\frac{(n^2x^2-2)\cos nx}{n^3}\\ \int x^2\cos(nx)dx&=\frac{2x\cos nx}{n^2}+\frac{(n^2x^2-2)\sin nx}{n^3}\\ \end{align}

Answer 2

Note that $f(x) = (\pi^2-x^2)\sin(nx)$ is an odd function, i.e., we have $f(-x) = -f(x)$. Hence, $$\int_{-a}^a f(x) dx = 0$$