I think you'll find that and ...
Given that n is a +ve odd integer:
Find f(x) = integral (from pi/2 to 0) of cos(x) * sin ^n/2 (x) dx
Using substitution u = sin x, i can arrive at:
F(X) = 2/(n+2) * [sin(x)] ^(n/2+1), from pi/2 to 0
however the answer states it as 2/(n+2), so my question is,
is there something i am missing out? Since [sin(x)] ^(n/2+1) at x = pi/2 will always give a decimal, as n is odd, so this term can never be cancelled out?
Thanks.