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

Theorem

$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$

Proof

The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$

Basis for the Induction

$\map P 1$ is the case:

\(\ds \int_0^{\frac \pi 2} \cos^2 x \rd x\)

\(=\)

\(\ds \frac \pi 4\)

Definite Integral from 0 to Half Pi of Square of Cosine x

\(\ds \)

\(=\)

\(\ds \dfrac 1 2 \times \dfrac \pi 2\)

\(\ds \)

\(=\)

\(\ds \dfrac 2 {\paren {2 \times 1}^2} \dfrac \pi 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 \times 1}!} {\paren {2^1 \times 1!}^2} \dfrac \pi 2\)

Thus $\map P 1$ 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\ds \int_0^{\frac \pi 2} \cos^{2 k} x \rd x = \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2$

from which it is to be shown that:

$\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} } x \rd x = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2$

Induction Step

This is the induction step:

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

\(\ds I_{k + 1}\)

\(=\)

\(\ds \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } I_k\)

Reduction Formula for Definite Integral of Power of Cosine

\(\ds \)

\(=\)

\(\ds \frac {2 \paren {k + 1} - 1} {2 \paren {k + 1} } \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds \frac {2 \paren {k + 1} \paren {2 \paren {k + 1} - 1} } {2^2 \paren {k + 1}^2} \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2\)

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

Therefore:

$\ds \forall n \in \Z_{> 0}: \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$

$\blacksquare$