# Trigonometric Integral excersice

1. Feb 28, 2007

### alba_ei

1. The problem statement, all variables and given/known data
Integral of $$\int \sin^{11/3}\alpha\, d\alpha$$

2. Relevant equations
$$\sin^2\alpha = 1 - cos^2\alpha$$

3. 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 dont know what else to do. any hints or tips?

Last edited: Feb 28, 2007
2. Feb 28, 2007

### dextercioby

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

3. Feb 28, 2007

### 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.

4. Feb 28, 2007

### alba_ei

It doesnt have upper or lower limits im just looking for the antiderivate

5. Feb 28, 2007

### dextercioby

There you go: add a constant to the result.

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