Definite Integral from 0 to Half Pi of Odd Power of Cosine x/Proof 1

Proof
The proof proceeds by induction.

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

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}$

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

Induction Step
This is the induction step:

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

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$