# Integral help

## Main Question or Discussion Point

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

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
$\int - \cot x \csc x - \csc ^2 x dx$

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

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

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

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