indefinite trigonometric integral

$\displaystyle \int \frac {\cos^5 x}{sqrt(sin x)} dx$

I haven't been able to figure out a way to deal with the square root in the denominator.

The solution the book provides makes it clear that i should be using a Pythagorean identity in the usual way.

Re: indefinite trigonometric integral

Look at $\displaystyle \int \frac{(\cos^2(x))^2(\cos(x))}{\sqrt{\sin(x)}}dx$.

I get there by saying to myself: "I'd *like* to substitute $\displaystyle u = \sin(x)$ to simply that ugly denominator. That means I need $\displaystyle du = \cos(x)dx$. OK, I'll peel one of those cosines in the numerator off and put it with the $\displaystyle dx$. If I'm lucky, the rest of the numerator can be expressed in terms of u. Looking at it, I see that I am indeed lucky."