How do I prove the relationship between $I_n:=\int_{0}^{\pi}(\sin x)^ndx$ and $I_n=\frac{n-1}{n}I_{n-2}$ by partial Integration?

Question

For all $n \in \mathbb{N} : n≥2$, I might add. $$I_n:=\int_{0}^{\pi}(\sin x)^ndx$$ $$I_n=\frac{n-1}{n}I_{n-2}$$

I've tried to rewrite $\int(\sin x)^ndx$ to the form $\int(\sin x)(\sin x)^{n-1}dx$ but then I don't get much further. Any advice on how to move forward would be appreciated.

Answer 1

The initial step is correct; assuming $n>1$, \begin{align} I_n &=\int_0^\pi (\sin x)^{n-1}\sin x\,dx \\ &=\Bigl[-(\sin x)^{n-1}\cos x\Bigr]_0^\pi +(n-1)\int_0^\pi (\sin x)^{n-2}\cos^2x\,dx\\ &=(n-1)\int_0^\pi (\sin x)^{n-2}(1-\sin^2x)\,dx\\ &=(n-1)\int_0^\pi (\sin x)^{n-2}\,dx- (n-1)\int_0^\pi (\sin x)^{n}\,dx\\[6px] &=(n-1)I_{n-2}-(n-1)I_n \end{align}

Answer 2

$I_n= \int ^{\pi}_0 \sin^{n-1}x.\sin{x}$ $dx=\sin^{n-1}x.(-\cos{x})-\int ^{\pi}_0 (n-1)\sin^{n-2}x.\cos{x}(-\cos{x})$ $dx$

$=\sin^{n-1}x.(-\cos{x})+\int ^{\pi}_0 (n-1)\sin^{n-2}x.\cos^2{x}$ $dx$

$=\sin^{n-1}x.(-\cos{x})+\int ^{\pi}_0 (n-1)\sin^{n-2}x.(1-\sin^2{x})$ $dx$

$=\sin^{n-1}x.(-\cos{x})|_0^{\pi}+(n-1)\int ^{\pi}_0 (\sin^{n-2}x-\sin^n{x})$ $dx$

$2I_n=0+(n-1)\int ^{\pi}_0 (\sin^{n-2}x)$ $dx=(n-1)I_{n-2}$

$\therefore I_n=\frac{(n-1)}{2}I_{n-2}$