# 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

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

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