I-n-0-pi-2-sin-n-x-dx-Write-a-relation-between-I-n-2-and-I-n-

Question Number 137187 by mathocean1 last updated on 30/Mar/21

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

W r i t e a r e l a t i o n b e t w e e n I_{n + 2} a n d I_{n} .

Answered by Dwaipayan Shikari last updated on 30/Mar/21

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

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

J_{n + 2} = \frac{Γ (\frac{n + 3}{2}) Γ (\frac{1}{2})}{2 Γ (\frac{n}{2} + 2)} = \frac{(\frac{n + 1}{2})}{(\frac{n}{2} + 1)} J_{n}

Answered by mathmax by abdo last updated on 30/Mar/21

I_{n + 2} = \int_{0}^{\frac{π}{2}} \sin^{n} x \sin^{2} xdx = \int_{0}^{\frac{π}{2}} \sin^{n} x (1 - \cos^{2} x) dx

= \int_{0}^{\frac{π}{2}} \sin^{n} xdx - \int_{0}^{\frac{π}{2}} \cos^{2} x \sin^{n} x dx = I_{n} - J

J = \int_{0}^{\frac{π}{2}} cosx (cosx \sin^{n} x) dx =_{by parts} {[\frac{\sin^{n + 1} x}{n + 1} cosx]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (- sinx) \frac{\sin^{n + 1} x}{n + 1} dx

= \frac{1}{n + 1} \int_{0}^{\frac{π}{2}} \sin^{n + 2} xdx \Rightarrow I_{n + 2} = I_{n} - \frac{1}{n + 1} I_{n + 2} \Rightarrow

(1 + \frac{1}{n + 1}) I_{n + 2} = I_{n} \Rightarrow \frac{n + 2}{n + 1} I_{n + 2} = I_{n} \Rightarrow I_{n + 2} = \frac{n + 1}{n + 2} I_{n}