How to compute $I(n)=\int_{0}^{\pi}x \sin^n(x)\,dx$?

Question

I would like to collect some hints for computing $$I(n)=\int_{0}^{\pi}x \sin^n(x)\,dx$$ for $n\in\mathbb{N}$.

Answer 1

The usual symmetry trick works just fine. $\sin(\pi -x)=\sin(x)$, hence:

$$ I(n) = \frac{1}{2}\left[\int_{0}^{\pi}(\pi-x)\sin^n(x)\,dx+\int_{0}^{\pi}x\sin^n(x)\,dx\right]=\pi \int_{0}^{\pi/2}\sin^n(x)\,dx $$ but: $$\begin{eqnarray*} \int_{0}^{\pi/2}\sin^n(x)\,dx &=& \int_{0}^{1} u^n(1-u^2)^{-1/2}\,dx\\&=&\frac{1}{2}\int_{0}^{1}v^{(n-1)/2}(1-v)^{-1/2}\,dv\\&=&\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\,\Gamma\left(\frac{n+2}{2}\right)}\end{eqnarray*} $$ so:

$$ I(n) = \color{red}{\frac{\pi\sqrt{\pi}}{2}\cdot \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}}.$$