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 $I_n$.

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

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

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

From Reduction Formula for Integral of Power of Sine:


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

Thus:

Also see

 * Reduction Formula for Definite Integral of Power of Cosine