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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.30$