Trigonometric Integral excersice

  • Thread starter alba_ei
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  • #1
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Homework Statement


Integral of [tex] \int \sin^{11/3}\alpha\, d\alpha [/tex]


Homework Equations


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


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?
 
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Answers and Replies

  • #2
dextercioby
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Interesting integral. Mathematica returns an answer involving the Gauss hypergeometric function [itex] _{2}F_{1} [/itex]
 
  • #3
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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.
 
  • #4
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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.

It doesnt have upper or lower limits im just looking for the antiderivate
 
  • #5
dextercioby
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There you go: add a constant to the result.
 

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  • http_integrals.wolfram.com_Integrator_MSP_MSPStoreID=M.pdf
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