the journal of francis billones

trivial math boredom

18 Dec 2023

Let’s say you wanted to compute for the following integral:

$$\begin{aligned} \int sin^Nx \, \mathrm{d}x \end{aligned}$$

where \(N = 2n + 1, n \in \mathbb{Z}\). In English, you want to get the integral of the sine function raised to an odd integer power.

First, we have to represent the integrand in a form wherein we can perform u-substitution. We can do so by separating one of the sines in the exponential:

$$\begin{aligned} \int \sin^{2n+1}x \, \mathrm{d}x = \int \sin^{2n}(x)\sin(x) \, \mathrm{d}x \end{aligned}$$

and then, using the identity \(\sin^{2}x = 1 — \cos^{2}x\):

$$\begin{aligned} \int \sin^{2n}(x)\sin(x) \, \mathrm{d}x = \int (\sin^{2}(x))^{n}\sin(x) \, \mathrm{d}x = \int (1-\cos^{2}(x))^{n}\sin(x) \, \mathrm{d}x \end{aligned}$$

Then, we can set \(u = \cos x, \mathrm{d}u = -\sin{x}\mathrm{d}x\):

$$\begin{aligned} \int (1-\cos^{2}(x))^{n}\sin(x) \, \mathrm{d}x = -\int (1-u^{2})^{n} \, \mathrm{d}u \end{aligned}$$

Now, you can stop here, but I didn’t. I wanted to get a closed form solution (or rather, as close to closed form as possible).

On the use of "closed form" I'm well aware that "closed form" is not the appropriate term here, but I wanted to use it to express the idea of a solution that is simply plug-and-play. Leaving the integral as is would require extensive work for N > 2. As for "extensive work", I mean something that requires a CAS to do more than just simple integer multiplication.


Conveniently enough, since \((1 — u^2)^n\) is a binomial, we can simply use the binomial theorem:

$$\begin{aligned} (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \end{aligned}$$

For our binomial \((1 — u^2)^n\), \(x\) would correspond to \(1\), and \(y\) would correspond to \(-u^2\):

$$\begin{aligned} (1 -u^{2})^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} (-u^2)^k \end{aligned}$$

which, when using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), as well as removing the redundant \(1^{n-k}\) (as 1 raised to any integer is 1), results in:

$$\begin{aligned} (1 -u^{2})^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} (-u^2)^k \end{aligned}$$

As this is simply the expansion of a binomial, which is a summation of terms each with a real coefficient and \(u\) raised to an integer power, when we substitute this back into the integral, we can simply use the power rule of integration:

$$\begin{aligned} \int \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} (-u^2)^k \, \mathrm{d}u = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!(2k+1)} (-u^2)^{k+1} \end{aligned}$$

and there we have it. An expression for \(\int sin^Nx \, \mathrm{d}x\) that requires nothing more than simple arithmetic.