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Great Fun: Complex Proof using de Moivre's Theorem

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that
    [tex]\int_{0}^{\pi} {\cos}^{2n}\theta d\theta = \frac{(2n)!\pi}{{2}^{2n}{(n!)}^{2}}[/tex]

    2. Relevant equations

    cos n*theta = (z^n + z^-n)/2

    3. The attempt at a solution

    Ok, this one has kept me busy for a while. I started using de Moivre's theorem to express cos^2n(theta) in terms of multiple angles but got stuck on the integral.
    After a while I tried to prove it inductively; showing it true for n=1 but then failed to show it for n=k+1.
    If someone could suggest a novel way to start this problem I would be very grateful.

  2. jcsd
  3. Nov 5, 2007 #2


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    1) First notice that the cosine is an even function, so if you integrate over -pi to +pi, the answer is double what you want.

    2) cos theta = (z + z^-1)/2, z=e^(i theta)

    3) What is the integral from -pi to +pi of z^n where n is an integer (possibly negative or zero)?
  4. Nov 6, 2007 #3
    Thanks for your reply Avodyne.

    To solve it, I expressed cos^2n(theta) as (z + z^-1)^2n. This expanded to yield:
    (cos(2n) + 2nC1*cos(2n-2) + ... + (2n)!/(n1)^2) / 2^2n

    When integrated, all the cos's become sin's. The sine of zero or any integer multiple of 2pi will cancell, therefore the only term remaining will be (2n)!/(n1)^2) / 2^2n.

    Retrospectively it seems rather easy, although it did keep me busy for a while.
  5. Nov 6, 2007 #4


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    You could also notice that [itex]\int_{-\pi}^{+\pi} e^{in\theta} d\theta[/itex] is zero unless n=0, in which case it is 2pi. This skips the step of assembling the z's into cosines.

    Problems are always easier in retrospect ...
  6. Nov 6, 2007 #5
    Yes they do :)
    Thanks again for the help.
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