1. Indefinite trig integral

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

I'm confused about the $sin^2x$ I tried the substitution $u = sinx \Rightarrow du = cosx dx$ which is in the integral.
but what does the new integral become?

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

Is the sub rule necessary for this integral?

Are there any trig identities which would help me out?

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

2. Re: Indefinite trig integral

You are correct with your substitution u=sinx so sin^2x=(sinx)^2=u^2

3. Re: Indefinite trig integral

ok, then the integral is $\int\frac{du}{u^2}$ which is $\frac{3}{u^3}+C$ correct?

4. Re: Indefinite trig integral

nvmnd did it with $\int u^{-1}du$

5. Re: Indefinite trig integral

Originally Posted by Foxlion
ok, then the integral is $\int\frac{du}{u^2}$ which is $\frac{3}{u^3}+C$ correct?
$\int\frac{du}{u^2}=\frac{-1}{u}+C$

6. Re: Indefinite trig integral

thank you both