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.