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Trigonometric Integral excersice

  1. Feb 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Integral of [tex] \int \sin^{11/3}\alpha\, d\alpha [/tex]


    2. Relevant equations
    [tex]\sin^2\alpha = 1 - cos^2\alpha [/tex]


    3. The attempt at a solution
    [tex]\int (\sin^2\alpha)^{4/3}\sin\alpha \, d\alpha[/tex]

    [tex]\int (1-cos^2\alpha)^{4/3}\sin\alpha \, d\alpha[/tex]

    [tex]u = \cos\alphad[/tex]
    [tex]du = \sin\alpha\, d\alpha [/tex]

    [tex]\int (1-u^2)^{4/3} du [/tex]
    i dont know what else to do. any hints or tips?
     
    Last edited: Feb 28, 2007
  2. jcsd
  3. Feb 28, 2007 #2

    dextercioby

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    Interesting integral. Mathematica returns an answer involving the Gauss hypergeometric function [itex] _{2}F_{1} [/itex]
     
  4. Feb 28, 2007 #3
    the integral seems pretty hopeless to evaluate due to the cubic root... are there bounds to the integral? that could potentially simply things a whole lot.
     
  5. Feb 28, 2007 #4
    It doesnt have upper or lower limits im just looking for the antiderivate
     
  6. Feb 28, 2007 #5

    dextercioby

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    There you go: add a constant to the result.
     

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