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Extremely confusing integral

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the integral of

    [itex]\int\frac{cos(x)}{1+e^{x}}[/itex]

    2. Relevant equations

    Given that

    [itex]\frac{1}{1+e^{x}}[/itex]-[itex]\frac{1}{2}[/itex] is an odd function

    3. The attempt at a solution

    I tried integration by parts, with both u = cos(x) and u = 1+e^x, and both only complicated it even more. I've not learned infinite series and sequences so I'm not sure thats the way.

    I did try writing cos(x) as [itex]\frac{e^{ix}+e^{-ix}}{2}[/itex], but I'm not sure what to do after that. I can see that there'd be some hyperbolic stuff in here but again, I'm not sure where exactly to start.
     
  2. jcsd
  3. Oct 3, 2012 #2

    LCKurtz

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    I don't see how that is relevant unless you are doing a definite integral on an interval like [-a,a].
    I don't think there is an elementary antiderivative.
     
  4. Oct 3, 2012 #3
    Oh jeez I forgot to put the interval. its from -pi/2 to pi/2. I'm really sorry about that
     
  5. Oct 3, 2012 #4

    LCKurtz

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    If ##f(x)## is odd what do you know about ##\int_{-a}^a f(x)\, dx##?
     
  6. Oct 3, 2012 #5

    SammyS

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    So, what is [itex]\displaystyle
    \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?[/itex]
     
  7. Oct 3, 2012 #6
    It would equal 0. So as SammyS put it, it would be [itex]\displaystyle
    \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?[/itex], where the first term would go to zero and then I'm left to evaluate the last bit which would be -[itex]\int_{-\pi/2}^{\pi/2}\frac{cos(x)}{2}[/itex]?

    Is this just the integral of the right side then? So it's just -1?
     
  8. Oct 3, 2012 #7

    LCKurtz

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    No. You don't evaluate an integral in pieces like that. You have ##\frac{1}{1+e^{x}}-\frac{1}{2}## multiplied by ##\cos x##. Is ##\cos x ## even, odd, or neither? What about its product with ##\frac{1}{1+e^{x}}-\frac{1}{2}##?
     
  9. Oct 3, 2012 #8

    SammyS

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    Remember, the hint says that [itex]\displaystyle \frac{1}{1+e^{x}}-\frac{1}{2}\ [/itex] is odd (that entire expression).

    How do you get [itex]\displaystyle \frac{1}{1+e^{x}}-\frac{1}{2}\ [/itex] from [itex]\displaystyle \frac{1}{1+e^{x}}\ ?[/itex]
     
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