prove-that-0-pi-2-sin-2k-x-dx-2k-2-2k-1-k-2-

Question Number 138112 by 676597498 last updated on 10/Apr/21

p r o v e t h a t

\int_{0}^{\frac{π}{2}} ({s i n}^{2 k} (x) d x) = \frac{(2 k)! 2^{- 2 k - 1}}{{(k!)}^{2}}

Answered by Dwaipayan Shikari last updated on 10/Apr/21

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

M a t c h i n g t h i s

\int_{0}^{\frac{π}{2}} {s i n}^{2 k} (x) d x = \int_{0}^{\frac{π}{2}} {s i n}^{2 (k + 1) - 1} x {c o s}^{2 (\frac{1}{2}) - 1} x = \frac{Γ (k + 1) Γ (\frac{1}{2})}{Γ (k + \frac{3}{2})}

Answered by mathmax by abdo last updated on 10/Apr/21

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

= \int_{0}^{\frac{π}{2}} \sin^{n - 2} xdx - \int_{0}^{\frac{π}{2}} \cos^{2} x \sin^{n - 2} xdx we [have

\int_{0}^{\frac{π}{2}} \sin^{n - 2} xdx = I_{n - 2} and by recurrence

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

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

\Rightarrow \prod_{k = 1}^{n} I_{2 k} = \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} I_{2 k - 2} \Rightarrow

I_{2} . I_{4} \dots \dots I_{2 n} = \frac{\prod_{k = 1}^{n} (2 k - 1)}{2^{n} n!} I_{0} . I_{2} \dots . . I_{2 n - 2} \Rightarrow

I_{2 n} = \frac{1.3.5 \dots . (2 n - 1)}{2^{n} . n!} I_{0} = \frac{π}{2} \frac{1.2.3.4.5 \dots . . (2 n - 1) (2 n)}{2^{n} n! . 2^{n .} n!}

= \frac{π}{2} \times \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \Rightarrow \int_{0}^{\frac{π}{2}} \sin^{2 n} xdx = \frac{π (2 n)!}{2^{2 n + 1} {(n!)}^{2}}