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

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Theorem

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


Then:

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


Proof 1

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:

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

$\blacksquare$


Proof 2

\(\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x\) \(=\) \(\ds \int_0^{\frac \pi 2} \paren {\sin x}^{\frac 2 2 - 1} \paren {\cos x}^{2 \paren {n + \frac 1 2} - 1} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \Beta \paren {\frac 1 2, n + \frac 1 2}\) Definition 2 of Beta Function
\(\ds \) \(=\) \(\ds \frac 1 2 \cdot \frac {\map \Gamma {n + \frac 1 2} \, \map \Gamma {\frac 1 2} } {\map \Gamma {n + 1} }\) Definition 3 of Beta Function
\(\ds \) \(=\) \(\ds \frac {\map \Gamma {n + \frac 1 2} \sqrt \pi} {2 \paren {n!} }\) Gamma Function of One Half
\(\ds \) \(=\) \(\ds \frac {\paren {2 n}! \paren {\sqrt \pi}^2} {2 \cdot 2^{2 n} \paren {n!}^2}\) Gamma Function of Positive Half-Integer
\(\ds \) \(=\) \(\ds \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2\)

$\blacksquare$


Sources

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