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

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Theorem

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


Then:

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


Proof 1

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:

\(\ds \int_0^{\frac \pi 2} \cos x \rd x\) \(=\) \(\ds \bigintlimits {\sin x} 0 {\frac \pi 2}\) Primitive of $\cos x$
\(\ds \) \(=\) \(\ds \sin \frac \pi 2 - \sin 0\)
\(\ds \) \(=\) \(\ds 1\) Sine of Right Angle, Sine of Zero is Zero
\(\ds \) \(=\) \(\ds \frac {\paren {2^0 \times 1!}^2} {\paren {2 \times 0 + 1}!}\)

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


\(\ds I_{k + 1}\) \(=\) \(\ds \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} I_k\) Reduction Formula for Definite Integral of Power of Cosine
\(\ds \) \(=\) \(\ds \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {\paren {2 \paren {k + 1} }^2} {\paren {2 \paren {k + 1} + 1} \paren {2 \paren {k + 1} } } \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}\) multiplying top and bottom by $2 \paren {k + 1}$
\(\ds \) \(=\) \(\ds \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}\)


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


Therefore:

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Sources

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