Integral help

1. Jan 28, 2007

bolas

I need help with this one:

$$\frac{1}{\cos(x)-1}dx$$

I know there must be a way to rewrite it so that i can use 1 - cos^2 = sin^2. I've tried to multiply by cosx + 1 / cosx + 1 and that gets me further but the answer doesn't seem to be the same maybe because of simplification?

Any hints?

thanks

2. Jan 28, 2007

IMDerek

The integral is equal to:

$\int \frac {\cos x + 1}{-\sin ^2 x} dx$ I suggest you use some trigonometric identities to change it into
$\int - \cot x \csc x - \csc ^2 x dx$

3. Jan 29, 2007

dextercioby

$$\int \frac{dx}{\cos x-1}=-\int \frac{dx}{1-\cos x}=-\frac{1}{2}\int \frac{dx}{\sin^{2}\frac{x}{2}}$$

use a simple substitution and a tabulated integral to get the answer.

4. Feb 5, 2007

reidy

$$\\T=[\sqrt\frac{\a}{g}]\In\fract{a+\sqrt{a^2-b^2}}{b}$$

Last edited: Feb 5, 2007
5. Feb 5, 2007

reidy

$$\\T={\sqrt\frac{\a}{g}}\In\fract{a+\sqrt{a^2-b^2}}{b}$$