# Thread: cosx/sin^2x - bit stuck

1. ## cosx/sin^2x - bit stuck

I can do these now but I am stuck on this indefinite integration.

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

So I thought try letting u = sinx. du/dx = cosx. And then.

$\int \frac{1}{u^2} du$

$= 2 ln |sinx| + c$

Angus

2. ## Re: cosx/sin^2x - bit stuck

Originally Posted by angypangy
I can do these now but I am stuck on this indefinite integration.
$\int \frac{cos x}{sin^2 x} dx$
So I thought try letting u = sinx. du/dx = cosx. And then.
$\int \frac{1}{u^2} du$
$= 2 ln |sinx| + c$
What is the derivative of $\frac{-1}{u}~?$

3. ## Re: cosx/sin^2x - bit stuck

Originally Posted by Plato
What is the derivative of $\frac{-1}{u}~?$
$\frac{-1}{u} = -u^{-1}$ so it would be $u^{-2}$ or $\frac{1}{u^2}$ Is that correct?

4. ## Re: cosx/sin^2x - bit stuck

Originally Posted by angypangy
$\frac{-1}{u} = -u^{-1}$ so it would be $u^{-2}$ or $\frac{1}{u^2}$ Is that correct?
Yes it is. Now look at what you said the antiderivative was.

5. ## Re: cosx/sin^2x - bit stuck

Originally Posted by Plato
Yes it is. Now look at what you said the antiderivative was.
So, just because the integral of 1/x is lnx, the integral of 1/u^2 is NOT ln x^2?

6. ## Re: cosx/sin^2x - bit stuck

Originally Posted by angypangy
So, just because the integral of 1/x is lnx, the integral of 1/u^2 is NOT ln x^2?
$\int \frac{\cos x}{\sin^2 x} dx=\frac{-1}{\sin(x)}+C$

7. ## Re: cosx/sin^2x - bit stuck

knowing your trig identities and basic trig derivatives may help ...

$\int \frac{\cos{x}}{\sin^2{x}} \, dx = \int \frac{1}{\sin{x}} \cdot \frac{\cos{x}}{\sin{x}} \, dx = \int\csc{x} \cdot \cot{x} \, dx = -\csc{x} + C$