How to calculate the Fourier Coefficient of $\sin^5(x)$ over $[-\pi,\pi]$?

Question

I would have to integrate $\sin^5(x)\cdot\sin(nx)$, but I have no idea how to. And that's the only coefficient I need for the series.

Answer 1

HINT

Use the half-angle formulae

$$\sin^2\left(\frac x2\right)=\frac{1-\cos(x)}2~~\text{and}~~\cos^2\left(\frac x2\right)=\frac{1+\cos(x)}2$$

Combined with the identities

$$\sin(a)\sin(b)=\frac12(\cos(a-b)-\cos(a+b))~~\text{and}~~\cos(a)\cos(b)=\frac12(\cos(a-b)+\cos(a+b))$$

Start with

$$\int_{-\pi}^\pi\sin^5(x)\sin(nx)\mathrm dx=\int_{-\pi}^\pi\left(\frac{1-\cos(2x)}2\right)^2\sin(x)\sin(nx)\mathrm dx=\dots$$

Can you take it from here?

Answer 2

Write $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$. Then raise to the fifth power and expand using the Binomial Theorem. You will find, using the fact that $(e^{x})^{n}=e^{nx}$ and cancelling appropriate terms, an expression for $\sin^{5}{x}$in terms of trigonometric functions of the form $\sin{nx},\cos{nx}$ ,which is exactly the Fourier expansion of your function.