Question involving integration and cosine

[pc2-brazil]

Homework Statement

Show that, if n is an odd number, \int_0^\pi \cos^nx dx = 0

Homework Equations

The Attempt at a Solution

\int_0^\pi \cos^nx dx = \int_0^\pi \cos^{n-1}(x)\cos (x) dx =
= \int_0^\pi (\cos^2x)^{\frac{n-1}{2}} \cos x dx = \int_0^\pi (1 - \sin^2x)^{\frac{n-1}{2}} \cos x dx
Now the next step would be to expand the term (1 - \sin^2x)^{\frac{n-1}{2}}. Then, I would be able to use u = sin(x) and du = cos(x) dx to eliminate the term cos(x).
It seems to make sense to expand (1 - \sin^2x)^{\frac{n-1}{2}} with the binomial theorem, but it would get very complicated. Is there a better way?

Thank you in advance.

 

[Dick]

Look for relation between the integral from 0 to pi/2 and the integral from pi/2 to pi. Sketch a graph.

 

[pc2-brazil]

Look for relation between the integral from 0 to pi/2 and the integral from pi/2 to pi. Sketch a graph.

I see what you mean. For the function y = cos x, the integral from π/2 to π is minus the integral from 0 to π/2 (because \cos(\frac{\pi}{2} + x) = -\cos(\frac{\pi}{2} - x), if x is a real number and 0\leq x\leq\frac{\pi}{2}). So, the integral from 0 to π is zero.
For the function cosⁿx, as long as n is odd, this relation will be maintained.
Is this correct? Is there a more formal way of showing it?

 

[Dick]

I see what you mean. For the function y = cos x, the integral from π/2 to π is minus the integral from 0 to π/2. So, the integral from 0 to π is zero.
For the function cosⁿx, as long as n is odd, this relation will be maintained.
But is there a more formal way of showing it?

Sure, split it into two integrals and do a change of variable like x=pi-u on the second one. Use some trig, like cos(pi-u)=(-1)*cos(u).