- #1
emjay66
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Homework Statement
I've been working on a problem from Apostol "Calculus" Volume 1 (not homework but self study). The problem is Section 5.8, Number 25 (Page 217) and states:
If [tex]$\m$[\tex] is a positive integer, show that:
[itex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx[/itex]
Homework Equations
The Attempt at a Solution
My thinking (so far) has been that I have two consider two cases,
- m odd
- m even
I use the substitution $$u = cos x, du = -sin x dx $
[itex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx =
- \int_0^{\frac{\pi}{2}} cos^m x sin^{m-1} x (-sin x) dx
= -\int_0^{\frac{\pi}{2}}cos^m x (sin^2 x)^{\frac{m-1}{2}}(-sin x) dx
= -\int_0^{\frac{\pi}{2}}cos^m x (1 - cos^2 x)^{\frac{m-1}{2}}(-sin x) dx
= -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du
= \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du
[/itex]
Since this is clearly a binomial expansion within the integral, it seems to me that I am moving away from the answer rather than towards the answer. I know that if $$m$$ odd, I would need to use the half angle formulas for $$sin x$$ and $$ cos x$$. In my opinion, I am clearly missing the point, so if anyone can point me in the right direction, that would be appreciated.