# Reduction Formula for Definite Integral of Power of Sine

## Theorem

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

Let $I_n$ be defined as:

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

Then $\left\langle{I_n}\right\rangle$ is a decreasing sequence of real numbers which satisfies:

$n I_n = \left({n - 1}\right) I_{n - 2}$

Thus:

$I_n = \dfrac {n - 1} n I_{n - 2}$

is a reduction formula for $I_n$.

### Corollary 1

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

### Corollary 2

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

## Proof

$\forall x \in \left[{0 \,.\,.\, \dfrac \pi 2}\right]: 0 \le \sin x \le 1$

and so on the same interval:

$0 \le \sin^{n + 1} x \le \sin^n x$

therefore:

$\forall n \in \N: 0 < I_{n + 1} < I_n$
$\displaystyle \int \sin^n x \rd x = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n$

Thus:

 $\displaystyle I_n$ $=$ $\displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$ $\displaystyle$ $=$ $\displaystyle \dfrac {n - 1} n \int_0^{\frac \pi 2} \sin^{n - 2} x \rd x - \left[{\dfrac {\sin^{n - 1} x \cos x} n}\right]_0^{\pi / 2}$ $\displaystyle$ $=$ $\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({\sin^{n - 1} \frac \pi 2 \cos \frac \pi 2 - \sin^{n - 1} 0 \cos 0}\right)} n$ $\displaystyle$ $=$ $\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - \sin^{n - 1} 0 \cos 0}\right)} n$ Cosine of Right Angle $\displaystyle$ $=$ $\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - 0}\right)} n$ Sine of Zero is Zero $\displaystyle \implies \ \$ $\displaystyle n I_n$ $=$ $\displaystyle \left({n - 1}\right) I_{n - 2}$ multiplying both sides by $n$

$\blacksquare$