# Definite Integral from 0 to Half Pi of Odd Power of Sine x

## Theorem

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

Then:

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

## Proof 1

Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$.

Then:

 $\displaystyle I_{2 n + 1}$ $=$ $\displaystyle \frac {2 n} {2 n + 1} I_{2 n - 1}$ Reduction Formula for Definite Integral of Power of Sine $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} } {\paren {2 n + 1} \paren {2 n - 1} } I_{2 n - 3}$ Reduction Formula for Definite Integral of Power of Sine again $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} I_1$ Reduction Formula for Definite Integral of Power of Sine until the end $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \int_0^{\pi / 2} \sin x \rd x$ Definition of $I_n$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \intlimits {-\cos x} 0 {\pi / 2}$ Primitive of Sine Function $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \paren {0 - \paren {-1} }$ Cosine of Right Angle and Cosine of Zero is One $\displaystyle$ $=$ $\displaystyle \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 n}^2 \paren {2 n - 2}^2 \dotsm 2^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}$ multiplying top and bottom by top $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 n}^2 n^2 \paren {n - 1}^2 \dotsm 1^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}$ extracting factor of $\left({2^n}\right)^2$ from the top $\displaystyle$ $=$ $\displaystyle \frac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$ Definition of Factorial

$\blacksquare$

## Proof 2

 $\displaystyle \int_0^{\pi/2} \sin^{2 n + 1} x \rd x$ $=$ $\displaystyle \int_0^{\pi/2} \paren {\sin x}^{2 \paren {n + 1} - 1} \paren {\cos x}^{2 \paren {\frac 1 2} - 1} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 \map \Beta {n + 1, \frac 1 2}$ Definition 2 of Beta Function $\displaystyle$ $=$ $\displaystyle \frac {\map \Gamma {n + 1} \map \Gamma {\frac 1 2} } {2 \map \Gamma {n + \frac 3 2} }$ Definition 3 of Beta Function $\displaystyle$ $=$ $\displaystyle \frac {n! \sqrt \pi} {2 \paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} }$ Gamma Function Extends Factorial, Gamma Function of One Half, Gamma Difference Equation $\displaystyle$ $=$ $\displaystyle \frac {n! \sqrt \pi} {2 n + 1} \times \frac {2^{2 n} n!} {\paren {2 n!} \sqrt \pi}$ Gamma Function of Positive Half-Integer $\displaystyle$ $=$ $\displaystyle \frac {\paren {2^n n!}^2 } {\paren {2 n + 1}!}$

$\blacksquare$