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Integral of (secx)^3 with eulers formula

  1. Jun 25, 2009 #1
    is it possible to integrate (secx)^3 with eulers formula
    could we use that cosx = (e^(ix) + e^(-ix)) /(2)
    then take it to the -3 power and multiply it out and try to integrate sec(x)^3 this way.
    this is not a homework ?
     
  2. jcsd
  3. Jun 25, 2009 #2
    This will only work well if you integrate from zero to 2 pi. But in that case this particular integral will be divergent.


    A rational function of cos and sin integrated from zero to 2 pi amounts to a contour integral of a rational function over the unit circle in the complex plane, so you can directly apply the residue theorem.
     
  4. Jun 26, 2009 #3
    can u give me an idea on how to start to integrate this.
     
  5. Jun 26, 2009 #4

    HallsofIvy

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    It's pretty much just algebra, isn't it?

    [tex]sec(x)= \frac{2}{e^x+ e^{-x}}[/tex]
    so
    [tex]sec^3(x)= \frac{8}{(e^x+ e^{-x})^3}[/tex]
    You can multiply both numerator and denominator by e3x to get
    [tex]\frac{8e^{3x}}{(e^x(e^x+ e^{-x}))^3}= \frac{8e^{3x}}{(e^{2x}+ 1)^3}[/tex]
    and your integral becomes
    [tex]\int\frac{8e^{3x}dx}{(e^2x+ 1)^3}[/tex]

    If you let u= ex, du= exdx and we have
    [tex]\int\frac{8u^2 du}{(u^2+ 1)^3}[/tex]
    which can be done in terms of partial fractions.
     
  6. Jun 26, 2009 #5
    thanks for doing this it must have taken you a long time ,
    But when say multiply both top and bottom by e^(3x)
    do you mean e^(3ix) or e(3x)
     
  7. Jun 26, 2009 #6
    okay i got it now thanks
     
  8. Jun 26, 2009 #7

    Pengwuino

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    Yes it is with the imaginary exponentials. Simply replace everything with i3x and it should still follow
     
  9. Jun 27, 2009 #8
    Eh, seems kind of ugly. This integral has a very natural integration by parts solution.
     
  10. Jun 27, 2009 #9
    i wouldn't say very natural my whole goal was to find an easier way then by parts , but i think parts is easier
    but in the case of like (e^x)sinx dx this is easier with eulers formula then by parts.
     
  11. Jun 27, 2009 #10
    Well I meant it was natural in the sense that sec^2(x) is the derivative of tan(x) and sec(x) differentiated gives sec(x)tan(x) and that really lends itself to a clean solution through integrating by parts.

    As for (e^x)sinx, I would agree.
     
  12. Jun 27, 2009 #11
    yes i agree. but i was hoping eulers formula would yield an easier soultion but appartenlty not.
     
  13. Jun 27, 2009 #12

    HallsofIvy

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    Sorry about dropping the "i" !
     
  14. Jun 27, 2009 #13
    its ok i got it now .
     
  15. Jun 28, 2009 #14
    how do i take the arctan(e^(ix)) how do i make it into the real part.
     
  16. Jun 28, 2009 #15
    The real part of arctan[exp(ix)] is pi/4 for real x.

    if f(z) is an analytic function such that for real z we have that f(z) is real, then:

    f*(z) = f(z*)

    The real part of f(z) is thus given by:

    Re[f(z)] = [f(z) + f*(z)]/2 = [f(z) + f(z*)]/2

    If we put z = exp(i x) for real x, then we have z* = 1/z, therefore:

    Re[arctan(z)] = 1/2 [arctan(z) + arctan(1/z)] = 1/2 pi/2 = pi/4


    The fact that

    arctan(z) + arctan(1/z) = pi/2

    for all z follows directly from the fact that for real z the above identity is valid using analytic continuation.
     
  17. Jun 28, 2009 #16
    i see thanks
     
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