solve-by-betta-function-0-2pi-sin-5-z-dz-

Question Number 156369 by tabata last updated on 10/Oct/21

s o l v e b y b e t t a f u n c t i o n \int_{0}^{2 π} {s i n}^{5} (z) d z

Commented by tabata last updated on 10/Oct/21

? ? ? ? ? ? ? ?

Commented by aliyn last updated on 10/Oct/21

? ? ? ? ? ?

Answered by Mathspace last updated on 11/Oct/21

\int_{0}^{2 π} {s i n}^{5} x d x

= \int_{0}^{π} {s i n}^{5} x d x + \int_{π}^{2 π} {s i n}^{5} x d x (\to x = π + u)

= \int_{0}^{π} {s i n}^{5} x d x + \int_{0}^{π} (- {s i n}^{5} u) d u

= 0 (n o n e e d f o r b e t t a \dots!)