$\displaystyle \int\frac{cos x}{sin^2x}dx$

I'm confused about the $\displaystyle sin^2x$ I tried the substitution $\displaystyle u = sinx \Rightarrow du = cosx dx$ which is in the integral.

but what does the new integral become?

I have $\displaystyle \int\frac{du}{u}$ although I feel it should be$\displaystyle \int\frac{du}{u^2}$ neither seem correct.

Is the sub rule necessary for this integral?

Are there any trig identities which would help me out?

Also what is $\displaystyle \frac{d}{dx}sin^2x$?