# 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:

$\displaystyle \int_0^{\frac \pi 2} \sin^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$

## Proof

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

Then:

 $\displaystyle I_{2 n + 1}$ $=$ $\displaystyle \frac {2 n} {2 n + 1} I_{2 n - 1}$ $\quad$ Reduction Formula for Definite Integral of Power of Sine $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right)} {\left({2 n + 1}\right) \left({2 n - 1}\right)} I_{2 n - 3}$ $\quad$ Reduction Formula for Definite Integral of Power of Sine again $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right) \cdots 2} {\left({2 n + 1}\right) \left({2 n - 1}\right) \cdots 3} I_1$ $\quad$ Reduction Formula for Definite Integral of Power of Sine until the end $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right) \cdots 2} {\left({2 n + 1}\right) \left({2 n - 1}\right) \cdots 3} \int_0^{\pi / 2} \sin x \rd x$ $\quad$ Definition of $I_n$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right) \cdots 2} {\left({2 n + 1}\right) \left({2 n - 1}\right) \cdots 3} \left[{-\cos x}\right]_0^{\pi / 2}$ $\quad$ Primitive of Sine Function $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right) \cdots 2} {\left({2 n + 1}\right) \left({2 n - 1}\right) \cdots 3} \left({0 - \left({-1}\right)}\right)$ $\quad$ Cosine of Right Angle and Cosine of Zero is One $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2 n \left({2 n - 2}\right) \cdots 2} {\left({2 n + 1}\right) \left({2 n - 1}\right) \cdots 3}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\left({2 n}\right)^2 \left({2 n - 2}\right)^2 \cdots 2^2} {\left({2 n + 1}\right) \left({2 n}\right)\left({2 n - 1}\right) \left({2 n - 2}\right) \left({2 n - 3}\right) \cdots 3 \cdot 2}$ $\quad$ multiplying top and bottom by top $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\left({2^n}\right)^2 n^2 \left({n - 1}\right)^2 \cdots 1^2} {\left({2 n + 1}\right) \left({2 n}\right)\left({2 n - 1}\right) \left({2 n - 2}\right) \left({2 n - 3}\right) \cdots 3 \cdot 2}$ $\quad$ extracting factor of $\left({2^n}\right)^2$ from the top $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$ $\quad$ Definition of Factorial $\quad$

$\blacksquare$