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Trig integral question

  1. Nov 11, 2008 #1
    Hello all - first time posting to this forum and glad to find this place. Not really sure how to make the mathematical script appear so apologies for using regular txt.

    I've been trying to find the correct integral to the following equation.

    F(time) = (Wavelength/4) x cos (2 x pi x frequency x time)

    The integral I've come up with using the "integration by parts" rule follows:

    Integral F(t) = wavelength x sin(2 x pi x frequency x time) + C
    8 x pi x frequency

    To check the calculations, I put together a spread sheet using averages for the y value to measure areas under the curve in the first equation vs plugging in values for the second. The differences are on the order of 3 magnitude leading me to believe the calculation of the integral is not correct. (The formatting doesnt come out correct in the post. The denominator is supposed to diving in the wavelength -> time numerator portion of the equation.)

    Can anyone point out where I made a mistake? Thanks in advance.
  2. jcsd
  3. Nov 11, 2008 #2


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    Science Advisor
    Homework Helper

    Welcome to PF Boomer.
    I don't quite follow what you tried to do with a spreadsheet, but I can tell you that the integration is mathematically correct (as you can check by differentiating). Maybe your "averaging" procedure doesn't work as well as you think?
  4. Nov 11, 2008 #3
    [tex]\int \mathrm{dt} \, \mathrm{fkt}(t) = \int \mathrm{dt} \, \frac{\lambda}{4} \cos \left( 2\pi f t \right) = \frac{\lambda}{8 \pi f} \sin \left( 2\pi f t \right) + C \neq \lambda \sin \left( 2\pi f t \right) + C[/tex]
  5. Nov 11, 2008 #4

    king vitamin

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    Gold Member

    When you "plug in values" for the integral, are you taking differences (i.e. definite integrals intF(t2)-intF(t1))? Or have you calculated the constant based on initial conditions? If the differences of the areas you calculated from the areas and the integral all differ by +/- a constant, then you not only obtain confirmation of your data and calculation, but the integration constant for your integral.
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