I-n-0-pi-2-sin-n-x-dx-

Question Number 196928 by SANOGO last updated on 03/Sep/23

I_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n} x d x

Commented by Frix last updated on 03/Sep/23

\underset{0}{\int^{\frac{π}{2}}} \sin^{n} x d x = \frac{1}{2} B (\frac{1}{2}, \frac{n + 1}{2}) =

= \frac{\sqrt{π} Γ (\frac{n + 1}{2})}{2 Γ (\frac{n}{2} + 1)}

Facebook Tweet Pin