# Proving an integral identity similar to the beta function, but without using the beta function

I'm trying to calculate the volume of an $$n$$-dimensional hypersphere. The text I'm working out of breaks down the calculatin into a few different steps, and I'm stuck on the following one:

By differentiating the function $$\sin^n r \cos r$$, prove that $$\sigma_{n+1} = \dfrac{n}{n+1}\sigma_{n-1}$$, where $$\sigma_n = \displaystyle \int_0^\pi \sin^n r \, dr$$.

My question: Is there a way to do this without using the gamma or beta functions? After all, if the gamma and beta functions are fair game, the identity $$\sigma_{n+1} = \dfrac{n}{n+1}\sigma_{n-1}$$ is almost immediate depending on your definition of the beta function. Even if you want to re-derive the trigonometric definition of the beta function from scratch, you don't need to differentiate $$\sin^n r \cos r$$ to do this.

What I've tried: The derivative proposed is $$\dfrac{d}{dr} \sin^n r \cos r = n\sin^{n-1}r\cos^2 r - \sin^{n+1}r$$, and the integral of this from $$0$$ to $$\pi$$ is $$0$$, so from this we get $$\sigma_{n+1} = n\int_0^\pi \sin^{n-1}r\cos^2 r \, dr$$ So I'm done if I can show $$\displaystyle \int_0^\pi \sin^{n-1}r \cos^2 r \, dr = \frac{1}{n+1} \int_0^\pi \sin^{n-1}r \, dr$$.

I know how to solve this problem using the gamma function argument. But is there a way to avoid that for students less familiar with the gamma and beta functions? Is there a better way to prove this identity without relying on the gamma and beta functions?

Using $$\cos^2 r = 1 - \sin^2 r$$ in what you've already determined, you get
\begin{aligned} \sigma_{n+1} & = n\int_0^\pi \sin^{n-1}r\cos^2 r \, dr \\ & = n\int_0^\pi \sin^{n-1}r(1 - \sin^2 r) \, dr \\ & = n\int_0^\pi (\sin^{n-1}r - \sin^{n+1} r) \, dr \\ & = n\int_0^\pi \sin^{n-1}r \, dr - n\int_0^\pi \sin^{n+1}r \, dr \\ & = n\sigma_{n-1} - n\sigma_{n+1} \end{aligned}\tag{1}\label{eq1A}
\begin{aligned} (n+1)\sigma_{n+1} & = n\sigma_{n-1} \\ \sigma_{n+1} & = \frac{n}{n+1}\sigma_{n-1} \end{aligned}\tag{2}\label{eq2A}
• Wow. That was far simpler than I was making it... here I was thinking the $\frac 1{n+1}$ term would somehow materialize out of the right hand side. Thanks! May 16, 2020 at 5:38
• @DFord You're welcome. I've found when dealing with larger powers of $\sin$ and $\cos$ being integrated that the $\sin^2(x) + \cos^2(x) = 1$ identity is quite often useful. May 16, 2020 at 5:41