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Elliptic integrals

  1. Mar 25, 2005 #1
    The function of the type:

    [tex] \int {(x^2 + 1)^{5/2}}x dx [/tex]

    This is simple to integrate but the trigonometric function:

    [tex] \int 3/5{(\sec x)}^{5/3}x dx [/tex] is already a problem.

    The first gives:

    [tex] \int {(x^2 + 1)^{5/2}}x dx = \int {u}^{5/2}1/2 du = 1/2 \int u^{5/2} du = 1/2 ({2u^{7/2}/7 + C) = 1/7{(x^2 + 1)}^{7/2} + C [/tex]
     
    Last edited: Mar 25, 2005
  2. jcsd
  3. Mar 25, 2005 #2
    what are you trying to find out?
     
  4. Mar 26, 2005 #3

    dextercioby

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    Okay.There's no possible connection between the 2 integrals and the second is not an elliptical one.

    There's the result for

    [tex]\int x (\sec x)^{\frac{5}{3}} \ dx [/tex]


    Daniel.
     

    Attached Files:

  5. Mar 26, 2005 #4
    Thanks for the solution. I see that the 2nd is a generalized hypergeometric function. Is [tex] \int ( cos x )^{3/2} dx [/tex] also hypergeometric?
     
    Last edited: Mar 26, 2005
  6. Mar 26, 2005 #5

    dextercioby

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    Nope,that's elliptic.I think i've posted the solution in another thrread *looks for the solution*.Nope i confused it with another one.

    There it is

    [tex]\int \cos^{3/2}x \ dx [/tex]

    is equal to




    Daniel.
     

    Attached Files:

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