91Ó°ÊÓ

Recall the trigonometric identity for the double angle of cosine: $$ \cos{2x} = 2\cos^2{x} - 1. $$ We can rewrite this as $$ \cos^2{x} = \frac{1+\cos{2x}}{2}. $$

To find the integral, we'll use a recursive approach. Let \(I_p = \int_{0}^{2\pi} (\cos{x})^{2p} \, dx\). We'll calculate \(I_p\) in terms of \(I_{p-1}\) and so on, until we reach the base case, \(I_0\). First, let's rewrite \(I_p\) using the trigonometric identity: $$ I_p = \int_{0}^{2\pi} \left(\frac{1+\cos{2x}}{2}\right)^{p} \, dx. $$ Let's expand the expression inside the integral using the binomial theorem: $$ I_p = \int_{0}^{2\pi} \sum_{k=0}^{p}\binom{p}{k}\cos^{2k}{2x}\left(\frac{1}{2}\right)^{p} \, dx. $$ Now, we'll interchange the integral and the summation: $$ I_p = \sum_{k=0}^{p}\binom{p}{k}\left(\frac{1}{2}\right)^{p}\int_{0}^{2\pi}\cos^{2k}{2x} \, dx. $$ Since \(I_0 = \int_{0}^{2\pi} dx = 2\pi\), we will find \(I_p\) recursively for all \(p \in\mathbb{N}\).

We have found a recursive formula for \(I_p\), the integral of \((\cos{x})^{2p}\) over the interval from \(0\) to \(2\pi\): $$ I_p = \sum_{k=0}^{p}\binom{p}{k}\left(\frac{1}{2}\right)^{p}\int_{0}^{2\pi}\cos^{2k}{2x} \, dx, $$ with the base case \(I_0 = 2\pi\). This formula allows us to compute the integral for any \(p \in\mathbb{N}\).