0-pi-sinx-2n-dx-n-N-

Question Number 144174 by lapache last updated on 22/Jun/21

\int_{0}^{π} {(s i n x)}^{2 n} d x = \dots . ? \forall n \in N

Commented by Willson last updated on 22/Jun/21

4^{n} \sin^{2 n} t = C_{2 n}^{n} + 2 \underset{k = 0}{\sum^{n - 1}} {(- 1)}^{n} C_{2 n}^{n - k} \cos (2 kt)

4^{n} {\int_{0}}^{π} \sin^{2 n} t dt = π C_{2 n}^{n} + 2 \underset{0}{\underset{⏟}{\underset{k = 0}{\sum^{n - 1}} {(- 1)}^{k} {\int_{0}}^{π} \cos (2 kt) dt}}

\Rightarrow {\int_{0}}^{π} {(sint)}^{2 n} dt = π (\frac{C_{2 n}^{n}}{4^{n}})

Answered by Dwaipayan Shikari last updated on 22/Jun/21

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

x = z - \frac{π}{2} \Rightarrow \int_{0}^{\frac{π}{2}} {(- s i n (\frac{π}{2} - z))}^{2 n} + \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (x) d x

= \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (z) d z + \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (x) d x \overset{D u m m y v a r i a b l e}{=} 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (x) d x

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

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