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Integration proof

  1. Apr 15, 2013 #1
    I'm almost out of high school and have been trying to do some "harder" proofs, anyone, I'm not quite sure on how to proceed on this:

    prove that for all k:

    [tex] \displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0 [/tex]

    If anyone could start me off as I'll I have in my head is "try to express it into something you can integrate", but having no luck.
    Last edited: Apr 15, 2013
  2. jcsd
  3. Apr 15, 2013 #2


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    It's not clear what k is limited to. Is k all integers, all reals, etc? If k is an integer, will its range include negative values? What about k = 0?
  4. Apr 15, 2013 #3
    The parts before this question asked me to integrate cos^6(x) and sin^6x using identities from cos^6(x) - sin^6(x) and cos^6x + sin^6x which I done successfully. At the end of the question it says: "You might like to consider how you would prove that [tex] \displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0 [/tex] for all k, without having to derive a new identity for each value of k.
  5. Apr 15, 2013 #4


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    hint: use symmetry :wink:
  6. Apr 17, 2013 #5
    Could you expand on that? I tried rewriting cos in terms of sin, but still pretty stuck

  7. Apr 17, 2013 #6


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    cos2k(π/2 - x) - sin2k(π/2 - x) = … ? :wink:
  8. Apr 17, 2013 #7
    = [itex] sin^{2k}x - cos^{2k}x = -(cos^{2k}x - sin^{2k} x) [/itex]

    I feel so dumb for not getting this :(

    I tried for around 15 minutes playing around with what you gave me to show it's 0 but to no avail... I'll try again tomorrow and post if I've made progress. Thanks for your patience.
  9. Apr 17, 2013 #8


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    EDIT: Ignore my previous post, I'd made an error.
    Last edited: Apr 18, 2013
  10. Apr 17, 2013 #9

    Ray Vickson

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    Look at the graphs of sin(x) and cos(x). Can you get one of the graphs by shifting the other one to the right or the left?
  11. Apr 19, 2013 #10
    yes cos(x) = sin(x + π/2)
    also sin(x) = cos(x - π/2)

    I can see why the integral is 0, by drawing both of the graphs and shading in the required area, but I'm having real trouble formalising a proof to actually put onto paper :\
  12. Apr 19, 2013 #11


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    try substituting y = π - x, or y = π/2 - x (and dy = -dx), and seeing what it does to the integral of one of your shaded parts :smile:
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