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

Question Number 53694 by gunawan last updated on 25/Jan/19

\underset{0}{\int^{r π}} \sin^{2 n} x d x =

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

\int_{0}^{π} s i n x d x =∣ - c o s x ∣_{0}^{π} = - (c o s π - c o s 0) = 2

s o t h e a r e a o f e a c h l o o p o f f (x) = s i n x i s 2

f (x) = {s i n}^{2 n} x

f (r π - x) = {[s i n (r π - x)]}^{2 n} = [e i t t h e r + s i n x o r - s i n x]

b u t {[s i n (r π - x)]}^{2 n} = {s i n}^{2 n} x

n o w \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x w h e n f (2 a - x) = f (x)

s i n x = s i n (2 π + x) p e r i d o c i t y 2 π

\int_{0}^{r π} {s i n}^{2 n} x d x

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

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

= \frac{r}{2} \times 2 \int_{0}^{π} {s i n}^{2 n} x d x [\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x w h e n f (2 a - x) = f (x)]

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

g a m m a f u n c t i o n \dots

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

r \times 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 n + 1 - 1} {x c o s}^{2 \times \frac{1}{2} - 1} d x

= r \times \frac{⌈ (2 n + 1) ⌈ (\frac{1}{2})}{⌈ (2 n + 1 + \frac{1}{2})}

i h a v e t r i e d t o s o l v e . . o t h e r s p l s c h e c k \dots