Solving $\frac{1}{\cos(x)-1}dx$ - Any Hints?

[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

 

[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

 

[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.

 

[reidy]

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

 

[reidy]

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