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Definite Integral Proof

  1. Dec 27, 2011 #1
    1. The problem statement, all variables and given/known data
    I've been solving a problem, the solution is complete, however, I must prove that the following relation is true:

    [tex] \int_{0}^{\frac{\pi}{2}} sin^m 2x dx = \int_{0}^{\frac{\pi}{2}} cos^m x dx[/tex]
    for any m.
    2. Relevant equations

    -

    3. The attempt at a solution

    Well, I've trying to find some kind of solution by using substitution, however, when I computed both integrals in indefinite form on WolframAlpha, to see if I was following the right path, it showed me an answer that contained a "hypergeometric function", which I haven't learned yet.
     
  2. jcsd
  3. Dec 27, 2011 #2

    micromass

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    Try a suitable subsitution. How can you make a sine into a cosine??
     
  4. Dec 27, 2011 #3
    micromass:

    Using the Pythagorean trigonometric identity.
    Well, this would become (if I take the positive root): [tex] sin^m 2x = (1-cos^2 2x)^{\frac{m}{2}} [/tex]

    I'm trying 'u' = cos x this time.
     
  5. Dec 27, 2011 #4

    micromass

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    What is

    [tex]\sin(\frac{\pi}{2}-x)[/tex]

    ??
     
  6. Dec 27, 2011 #5
    micromass:

    cos x
     
  7. Dec 27, 2011 #6

    Curious3141

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    AndersCarlos, this is more of the same based on what we discussed yesterday.

    Micromass has given you a very big hint. Try to convert [itex]\sin 2x[/itex] into [itex]\cos u[/itex]. What substitution would do that? (Note that your sub must also convert that double angle into a single angle). Hint: There's a [itex]\frac{\pi}{4}[/itex] somewhere in there.

    After that, there's more of that "even function" manipulation we were talking about yesterday.
     
    Last edited: Dec 27, 2011
  8. Dec 27, 2011 #7
    micromass and Curious3141:

    Well, I chose that: 2x = π/2 - u
    then, dx = -du/2
    [tex]\int_{0}^{\frac{\pi}{2}} sin^m (2x)dx = - \int_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{sin^m (\frac{\pi}{2} - u)}{2}du = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{cos^m (u)}{2} du = \int_{0}^{\frac{\pi}{2}} cos^m (u) du = \int_{0}^{\frac{\pi}{2}} cos^m (x) dx [/tex]

    Well, I didn't see any π/4 during the process, but if there is anything wrong with this proof, sorry because I wrote it quite fast. Thank you both for your help.

    Edit: I have forgotten to put the 'm' exponent through the process, just fixed it.
     
    Last edited: Dec 27, 2011
  9. Dec 27, 2011 #8

    SammyS

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    Looks good !
     
  10. Dec 27, 2011 #9

    Curious3141

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    Well, the pi/4 is implicit in your proof. What's x in terms of u? :wink:

    Anyway, good job.:smile:
     
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