# Thread: Integral of sin^n (x)

1. ## Integral of sin^n (x)

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.

2. I think you'll find that $\displaystyle \sin{0} = 0$ and $\displaystyle \sin{\frac{\pi}{2}} = 1$...
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