Definite Integral from 0 to Half Pi of Even Power of Sine x

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Theorem

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


Then:

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


Proof

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


Then:

\(\displaystyle I_{2 n}\) \(=\) \(\displaystyle \frac {2 n - 1} {2 n} I_{2 n - 2}\) $\quad$ Reduction Formula for Definite Integral of Power of Sine $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n - 1}\right) \left({2 n - 3}\right)} {2 n \left({2 n - 2}\right)} I_{2 n - 4}\) $\quad$ Reduction Formula for Definite Integral of Power of Sine again $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n - 1}\right) \left({2 n - 3}\right) \cdots 1} {2 n \left({2 n - 2}\right) \cdots 2} I_0\) $\quad$ Reduction Formula for Definite Integral of Power of Sine until the end $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n - 1}\right) \left({2 n - 3}\right) \cdots 1} {2 n \left({2 n - 2}\right) \cdots 2} \int_0^{\pi / 2} \rd x\) $\quad$ Definition of $I_n$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n - 1}\right) \left({2 n - 3}\right) \cdots 1} {2 n \left({2 n - 2}\right) \cdots 2} \left[{x}\right]_0^{\pi / 2}\) $\quad$ Integral of Constant $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n - 1}\right) \left({2 n - 3}\right) \cdots 1} {2 n \left({2 n - 2}\right) \cdots 2} \frac \pi 2\) $\quad$ Integral of Constant $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2n \left({2 n - 1}\right) \left({2 n - 2}\right) \left({2 n - 3}\right) \cdots 2 \cdot 1} {\left({2 n}\right)^2 \left({2 n - 2}\right)^2 \cdots 2^2} \frac \pi 2\) $\quad$ multiplying top and bottom by bottom $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2n \left({2 n - 1}\right) \left({2 n - 2}\right) \left({2 n - 3}\right) \cdots 2 \cdot 1} {\left({2^n}\right)^2 n^2 \left({n - 1}\right)^2 \cdots 1^2} \frac \pi 2\) $\quad$ extracting factor of $\left({2^n}\right)^2$ from the bottom $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n}\right)!} {\left({2^n}\right)^2 \left({n!}\right)^2} \frac \pi 2\) $\quad$ Definition of Factorial $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 n}\right)!} {\left({2^n n!}\right)^2} \frac \pi 2\) $\quad$ rearranging $\quad$

$\blacksquare$


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