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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?

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Using $\cos^2 r = 1 - \sin^2 r$ in what you've already determined, you get

$$\begin{equation}\begin{aligned} \sigma_{n+1} & = n\int_0^\pi \sin^{n-1}r\cos^2 r \, dr \\ & = n\int_0^\pi \sin^{n-1}r(1 - \sin^2 r) \, dr \\ & = n\int_0^\pi (\sin^{n-1}r - \sin^{n+1} r) \, dr \\ & = n\int_0^\pi \sin^{n-1}r \, dr - n\int_0^\pi \sin^{n+1}r \, dr \\ & = n\sigma_{n-1} - n\sigma_{n+1} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This leads to

$$\begin{equation}\begin{aligned} (n+1)\sigma_{n+1} & = n\sigma_{n-1} \\ \sigma_{n+1} & = \frac{n}{n+1}\sigma_{n-1} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

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  • $\begingroup$ 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! $\endgroup$
    – D Ford
    May 16, 2020 at 5:38
  • $\begingroup$ @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. $\endgroup$
    – John Omielan
    May 16, 2020 at 5:41

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