Integration by substitution question

1. Mar 21, 2014

emjay66

1. The problem statement, all variables and given/known data
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 $$\m[\tex] is a positive integer, show that: $\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx$ 2. Relevant equations 3. The attempt at a solution My thinking (so far) has been that I have two consider two cases, 1. m odd 2. m even When I have tackled \m odd, the following occurred (and this is what confused me): I use the substitution u = cos x, du = -sin x dx  $\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = - \int_0^{\frac{\pi}{2}} cos^m x sin^{m-1} x (-sin x) dx = -\int_0^{\frac{\pi}{2}}cos^m x (sin^2 x)^{\frac{m-1}{2}}(-sin x) dx = -\int_0^{\frac{\pi}{2}}cos^m x (1 - cos^2 x)^{\frac{m-1}{2}}(-sin x) dx = -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du = \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du$ Since this is clearly a binomial expansion within the integral, it seems to me that I am moving away from the answer rather than towards the answer. I know that if m odd, I would need to use the half angle formulas for sin x and  cos x. In my opinion, Im clearly missing the point, so if anyone can point me in the right direction, that would be appreciated. 2. Mar 21, 2014 Simon Bridge Hmmm ... tidying that up first: [tex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx$$

2. Relevant equations

3. The attempt at a solution
My thinking (so far) has been that I have two consider two cases,
1. m odd
2. m even
When I have tackled $m$ odd, the following occurred (and this is what confused me):
I use the substitution $$u = \cos x, du = -\sin x dx$$

$$\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x dx \\ = - \int_0^{\frac{\pi}{2}} \cos^m x \sin^{m-1} x (-\sin x) dx\\ = -\int_0^{\frac{\pi}{2}}\cos^m x (\sin^2 x)^{\frac{m-1}{2}}(-\sin x) dx\\ = -\int_0^{\frac{\pi}{2}}\cos^m x (1 - \cos^2 x)^{\frac{m-1}{2}}(-\sin x) dx\\ = -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du\\ = \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du$$

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'm clearly missing the point, so if anyone can point me in the right direction, that would be appreciated.

3. Mar 21, 2014

Simon Bridge

I would have used the double-angle formulas myself - specifically: $$\sin 2x = 2\sin x\cos x$$
http://en.wikipedia.org/wiki/List_o...le.2C_triple-angle.2C_and_half-angle_formulae

There's a reduction formula for $\int \sin^n x\; dx$ and $\int \cos^n x\; dx$

You could also use the power reduction formulae on the integrands.
This divides into m odd vs m even.

Of course there is also: Note: $$\int \sin^nx\cos^nx\;\dx = -\frac{\sin^{n-1}x\cos^{n+1}x}{2n}+\frac{n-1}{2n}\int \sin^{n-2}x\cos^nx\;dx\; :\; n>0$$
http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions

You only need to show that the LHS = RHS after the definitite integral BTW: i.e. the areas must be the same, not the integrands.

Last edited: Mar 21, 2014
4. Mar 21, 2014

utkarshakash

I would suggest using Walli's Formula rather than doing the integration manually.