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Quite a complex integral

  1. Apr 29, 2009 #1
    Ok, I am trying to integrate the following function, and not getting very far: it's s=integral between 0 and 2pi of cos^-1(arctan((2*pi/b)*a*cos(2*pi*x/b)))dx)^-1 where a and b are known variables. What I would like to know, is can this integral be evaluated directly, or must I use the trapezium rule, and if so how would I do that?
     
  2. jcsd
  3. Apr 29, 2009 #2
    by known variables, do you mean constant in the formula? If so, it should be possible to integrate as what you then have is
    (cos-1(X*cos(2pi * x/b)-1dx
    to integrate. You should be able to integrate that although I can't remember what the integral of the inverse trig functions are off the top of my head. It may be a pretty complex integral but I'd expect it to be possible to do so with a closed-form solution.

    edit: Ignore that... I misread the original formula and didn't match up the brackets... need to re-evaluate
     
    Last edited: Apr 29, 2009
  4. Apr 29, 2009 #3
    yes, they are constants
     
  5. Apr 29, 2009 #4
    Just trying to visualize the formula.

    [tex]s= \displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx )^-1[/tex]

    What about the last missing ")^-1"?
     
  6. Apr 29, 2009 #5
    That is correct, and the whole integral is ^-1.
     
  7. Apr 29, 2009 #6
    so you would need a ( in front of the integral sign to complete it
     
  8. Apr 29, 2009 #7
    [tex]
    s= \dfrac{1}{\displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx}
    [/tex]

    I couldn't solve it. Neither did "Derive 6".

    Regards.
     
    Last edited: Apr 29, 2009
  9. Apr 29, 2009 #8
    yes, that is right, and a and b can be any numbers really, as they define the shape of the sine wave.
     
  10. Apr 29, 2009 #9

    MathematicalPhysicist

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    I haven't done any calculuations yet, but one thought that got into my attention, is perhaps you should take derivative under the integral sign of a or b, and use the theorem, that:
    (d/da)S f(x,a)dx=S df(x,a)/da dx.
    Not sure if that will ease the calculations, but it sure as hell worth the effort.

    Cheers.
     
  11. Apr 29, 2009 #10
  12. Apr 29, 2009 #11

    djeitnstine

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    Maxima got this from your basic function without the constants

    Code (Text):

    integrate(sec(atan(cos(x))), x);

    [tex]\int{\sqrt{cos^2(x)+1}}dx[/tex]
     
    which of course has no analytical solution =(

    edit: with a constant such as the 'a' here

    Code (Text):

    integrate(sec(atan(a*cos(x))), x);

    [tex]\int{\sqrt{a^2cos^2(x)+1}}dx[/tex]
     
     
  13. Apr 29, 2009 #12
    Last edited by a moderator: May 4, 2017
  14. Apr 29, 2009 #13
    This is a DEFINITE integral which means I'm sure that a computer program can make a numeric approximation, hopefully.
     
  15. Apr 29, 2009 #14

    djeitnstine

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    certainly approximations can be made but the OP would like to have an analytical answer.
     
  16. Apr 29, 2009 #15

    Mute

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    Waaaaaaiiiiiit just a minute here. The integral that GRfrones suggested,

    [tex]\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx [/tex]

    does not appear to be the integral in the original post (following the standard convention of notations):

    "integral between 0 and 2pi of cos^-1(arctan((2*pi/b)*a*cos(2*pi*x/b)))dx"

    GRfrones seems to have interpreted "cos^(-1)(stuff)" as "sec(stuff)", but I would take this to be "arccos(stuff)", as that's pretty much exclusively what "cos^(-1)(stuff)" means, and so I would have thought the integral is

    [tex]\int_0^{2\pi} \cos^{-1}\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx [/tex]

    Is it really GRfrones' interpretation? If so, then the OP should take care when writing "cos^(-1)" to mean "sec", as I can think of no instance where "cos^(-1)" would be taken to mean the secant.
     
  17. Apr 29, 2009 #16

    djeitnstine

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    The OP has already identified the correct integral
     
  18. May 1, 2009 #17

    Mute

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    I'm aware of that, I wanted to double-check, and point out that if that is the correct integral then the use of "cos^{-1}" to mean "sec" is not a good idea, as it conflicts with the standard usage of the symbol.
     
  19. May 2, 2009 #18
    Changing it from sec to arccos stops the integral from working on wolfram as well. I'm not sure that is a good or bad thing though, considering the integral it returns for the one identified as correct by the OP :)
     
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