Integration by substitution question

[emjay66]

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:<br /> <br /> $\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx$<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> My thinking (so far) has been that I have two consider two cases, <br /> <ol> <li data-xf-list-type="ol">m odd</li> <li data-xf-list-type="ol">m even</li> </ol>When I have tackled $$\m$$ odd, the following occurred (and this is what confused me):<br /> I use the substitution $$u = cos x, du = -sin x dx $<br /> $\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx =<br /> - \int_0^{\frac{\pi}{2}} cos^m x sin^{m-1} x (-sin x) dx<br /> = -\int_0^{\frac{\pi}{2}}cos^m x (sin^2 x)^{\frac{m-1}{2}}(-sin x) dx<br /> = -\int_0^{\frac{\pi}{2}}cos^m x (1 - cos^2 x)^{\frac{m-1}{2}}(-sin x) dx<br /> = -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du<br /> = \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du$<br /> Since this is clearly a binomial expansion within the integral, it seems to me that I am moving away from the answer rather than towards the answer. I know that if $$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.[/tex]

 

[Simon Bridge]

Hmmm ... tidying that up first:

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 $m$ is a positive integer, show that:

$$\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx$$

Homework Equations

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.

 

[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.

 

[utkarshakash]

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