1. ## Indefinite trig integral

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

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

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 $\displaystyle \int\frac{du}{u^2}$ which is $\displaystyle \frac{3}{u^3}+C$ correct?

4. ## Re: Indefinite trig integral

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

5. ## Re: Indefinite trig integral

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

6. ## Re: Indefinite trig integral

thank you both