consider-the-general-definite-intergral-I-n-0-pi-2-sin-n-xdx-a-prove-that-for-n-2-nI-n-n-1-I-n-2-b-Find-the-values-of-i-0-pi-2-sin-5-dx-ii-0-pi-2-sin-6-dx-

Question Number 63519 by Rio Michael last updated on 05/Jul/19

c o n s i d e r t h e g e n e r a l d e f i n i t e i n t e r g r a l

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

a) p r o v e t h a t f o r n ⩾ 2, {n I}_{n} = (n - 1) I_{n - 2} .

b) F i n d t h e v a l u e s o f i) \int_{0}^{\frac{π}{2}} {s i n}^{5} d x i i) \int_{0}^{\frac{π}{2}} {s i n}^{6} d x

Commented by Prithwish sen last updated on 05/Jul/19

I_{n} = \int_{0}^{\frac{π}{2}} \sin^{n - 1} x sinxdx, using by parts

Missing \left or extra \right

= (n - 1) I_{n - 2} - (n - 1) I_{n}

{nI}_{n} = (n - 1) I_{n - 2} proved

\int_{0}^{\frac{π}{2}} \sin^{5} x dx = \frac{(5 - 1)}{5} I_{3} = \frac{4}{5} . \frac{(3 - 1)}{3} I_{3 - 2} = \frac{4}{5} . \frac{2}{3} . I

= \frac{8}{15} \int_{0}^{\frac{π}{2}} sinx dx = - \frac{8}{15} {(cosx)}_{0}^{\frac{π}{2}} = \frac{8}{15}

similarly you can deduce (ii)

Commented by Rio Michael last updated on 05/Jul/19

s o i i) b e c o m e s \frac{5 π}{32} ?

Commented by Prithwish sen last updated on 05/Jul/19

yes

Commented by Rio Michael last updated on 05/Jul/19

o k a y s i r t h a n k s