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

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

Then:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ 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 $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\left({2^k k!}\right)^2} {\left({2 k + 1}\right)!}$

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

Induction Step
This is the induction step:

Let $I_k = \displaystyle \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x$.

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

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