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Question involving integration and cosine

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that, if n is an odd number, [itex]\int_0^\pi \cos^nx dx = 0[/itex]

    2. Relevant equations

    3. The attempt at a solution

    [tex]\int_0^\pi \cos^nx dx = \int_0^\pi \cos^{n-1}(x)\cos (x) dx = [/tex]
    [tex]= \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 [/tex]
    Now the next step would be to expand the term [itex](1 - \sin^2x)^{\frac{n-1}{2}}[/itex]. 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 [itex](1 - \sin^2x)^{\frac{n-1}{2}}[/itex] with the binomial theorem, but it would get very complicated. Is there a better way?

    Thank you in advance.
     
  2. jcsd
  3. Dec 1, 2011 #2

    Dick

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    Look for relation between the integral from 0 to pi/2 and the integral from pi/2 to pi. Sketch a graph.
     
  4. Dec 1, 2011 #3
    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 [itex]\cos(\frac{\pi}{2} + x) = -\cos(\frac{\pi}{2} - x)[/itex], if x is a real number and [itex]0\leq x\leq\frac{\pi}{2}[/itex]). 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?
     
    Last edited: Dec 1, 2011
  5. Dec 1, 2011 #4

    Dick

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    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).
     
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