Integral involving square root -

[winbacker]

Homework Statement

integrate sqrt(1-x^-2/3)^1/2.

Homework Equations

The Attempt at a Solution

The only thing I can think of is u substitution with u = 1 - x^-2/3. Obviously this cannot work because du differs by more than just a constant.

I guess I need to somehow factor this equation but I do not know how. I think I can pull out an x^1/3 or something but I'm not sure. Any help would be appreciated.

 

[maze]

Whenever you see stuff like that in the square root, always think "Trig Substitution"!

Hint: 1-sin(x)^2 = cos(x)^2.

 

[statdad]

Do you mean

<br /> \sqrt{1-x^{-2/3}}<br />

or do you mean (as you've written)
<br /> \sqrt{(1-x^{-2/3})^{1/2}}<br />

 

[winbacker]

the first one

 

[Mark44]

Whenever you see stuff like that in the square root, always think "Trig Substitution"!

Hint: 1-sin(x)^2 = cos(x)^2.

I'm not sure that trig substitution is the way to go in this problem. Trig substitution is a viable alternative for integrals that involve
\sqrt{a^2 + x^2}
\sqrt{a^2 - x^2}
\sqrt{x^2 - a^2}

Maybe it can be made to work in the OP's problem, but I don't see it.