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Integral of cosine function

  1. Jan 14, 2011 #1
    Hi. I have been experimenting a little to come up with the following "conjecture"
    [tex]
    \int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi=0
    [/tex]
    where a and b are arbitrary constants and f(x) is any function. Is this true? I guess it can be shown by expanding f in a power series of cosines?
     
  2. jcsd
  3. Jan 14, 2011 #2

    tiny-tim

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    hi daudaudaudau! :smile:

    doesn't work for f = √, a = b :wink:
     
  4. Jan 14, 2011 #3

    LeonhardEuler

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    Yes, I'm getting that that this is true. It can be proven using the substitution
    [tex]x=\cos{\phi}[/tex]
    [tex]dx=-\sin{\phi}d\phi[/tex]
    But this substitution is not 1-to-1: Each x value corresponds to 2 phi values on [0,2pi]. So you need to break the region of integration into [0,pi] and [pi,2pi]. If you look at the graph of the cos function, you will see it is 1-to-1 on these two intervals and goes from 1 to -1 on [0,pi] and from -1 to 1 on [pi,2pi]. So the integral becomes:
    [tex]
    \int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi= \int_1^{-1}-f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = \int_{-1}^{1}f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = 0 [/tex]


    Tiny-tim: I get this even in the example you gave.
     
  5. Jan 14, 2011 #4

    tiny-tim

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    hmm … i think i've been misled by the ambiguity of the √ function :redface:

    yes, we can get it directly from the original integral …

    if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]φ=0
     
  6. Jan 14, 2011 #5

    LeonhardEuler

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    Happens to everyone :wink:
    That is a better way of proving it.
     
  7. Jan 14, 2011 #6
    What happened to the sine function?
     
  8. Jan 14, 2011 #7

    tiny-tim

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    chain rule :wink:
     
  9. Jan 14, 2011 #8
    clever :-)
     
  10. Jan 14, 2011 #9

    AlephZero

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    The integral from -pi to pi is zero, because you are integrating an odd function.

    And the integral from -pi to pi equals the integral from 0 to 2pi.

    QED.
     
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