Definite Integral from 0 to Half Pi of Even Power of Cosine x

Theorem
Let $n \in \Z_{\ne 0}$ be a positive integer.

Then:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\left({2 n}\right)!} {\left({2^n n!}\right)^2} \dfrac \pi 2$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\left({2 n}\right)!} {\left({2^n n!}\right)^2} \dfrac \pi 2$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 k} x \rd x = \dfrac {\left({2 k}\right)!} {\left({2^k k!}\right)^2} \dfrac \pi 2$

from which it is to be shown that:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 \left({k + 1}\right)} x \rd x = \dfrac {\left({2 \left({k + 1}\right)}\right)!} {\left({2^{k + 1} \left({k + 1}\right)!}\right)^2} \dfrac \pi 2$

Induction Step
This is the induction step:

Let $I_k = \displaystyle \int_0^{\frac \pi 2} \cos^{2 k} x \rd x$.

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\left({2 n}\right)!} {\left({2^n n!}\right)^2} \dfrac \pi 2$