- #1

calisoca

- 28

- 0

## Homework Statement

I wasn't really sure where to post this, as I don't need help understanding the integration. I need help with the trigonometry!

That being said, here is the problem.

Find the general indefinite integral of [tex] \int\frac{\sin{x}}{1-sin^2{x}}dx [/tex]

## Homework Equations

[tex] \int\frac{\sin{x}}{1-sin^2{x}}dx [/tex]

## The Attempt at a Solution

I've tried starting from [tex] \cos^2{x} + \sin^2{x} = 1 [/tex].

That gets me to [tex] \frac{\sin{x}}{\cos^2{x}} [/tex], but I'm not sure what good that really does me?

I'm a little stuck here, and some help would be much appreciated!