Trigonometric Integral excersice

[alba_ei]

Homework Statement

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

Homework Equations

\sin^2\alpha = 1 - cos^2\alpha

The Attempt at a Solution

\int (\sin^2\alpha)^{4/3}\sin\alpha \, d\alpha

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

u = \cos\alphad
du = \sin\alpha\, d\alpha

\int (1-u^2)^{4/3} du
i don't know what else to do. any hints or tips?

 

[dextercioby]

Interesting integral. Mathematica returns an answer involving the Gauss hypergeometric function _{2}F_{1}

 

[tim_lou]

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.

 

[alba_ei]

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 doesn't have upper or lower limits I am just looking for the antiderivate

 

[dextercioby]

There you go: add a constant to the result.