Reduction Formula for Definite Integral of Power of Sine

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

Let $I_n$ be defined as:
 * $\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \ \mathrm d x$

Then $\left\langle{I_n}\right\rangle$ is a decreasing sequence of real numbers which satisfies:
 * $n I_n = \left({n - 1}\right) I_{n-2}$

is a reduction formula for $I_n$.

Proof
From Shape of Sine Function:
 * $\forall x \in \left({0 \,.\,.\, 1}\right): 0 < \sin x < \dfrac \pi 2$

and so on the same interval:
 * $0 < \sin^{n+1} x < \sin^n x$

therefore:
 * $\forall n \in \N: 0 < I_{n+1} < I_n$

From Reduction Formula for Integral of Power of Sine:


 * $\displaystyle \int \sin^n x \ \mathrm d x = \dfrac {n - 1} n \int \sin^{n - 2} x \ \mathrm d x - \dfrac {\sin^{n-1} x \cos x} n$

Thus: