0-pi-sin-2n-xdx-

Question Number 139883 by qaz last updated on 02/May/21

\int_{0}^{π} \sin^{2 n} x d x = ?

Answered by Dwaipayan Shikari last updated on 02/May/21

\int_{0}^{π} {s i n}^{2 n} (x) d x

= 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (x) d x = 2 . \frac{Γ (\frac{2 n + 1}{2}) Γ (\frac{1}{2})}{2 Γ (\frac{2 n}{2} + 1)} = \frac{Γ (n + \frac{1}{2}) Γ (\frac{1}{2})}{n!}

A s \int_{0}^{\frac{π}{2}} {s i n}^{a - 1} x {c o s}^{b - 1} x d x = \frac{Γ (a) Γ (b)}{2 Γ (a + b)}

Commented by Ar Brandon last updated on 02/May/21

= \frac{π n (2 n - 1)!}{2^{2 n - 1} {(n!)}^{2}}