# Integral with trig function in denominator

• Jun 25th 2012, 04:19 PM
Ragnarok
Integral with trig function in denominator
Another seemingly simple integral I am going blank on:

$\displaystyle \int \frac{1}{cosx-1}\,dx$

I really appreciate any help! I have tried different things but none of them seem to work.
• Jun 25th 2012, 04:34 PM
Reckoner
Re: Integral with trig function in denominator
$\displaystyle \int\frac1{\cos x - 1}\,dx$

$\displaystyle =\int\left(\frac1{\cos x - 1}\cdot\frac{\cos x + 1}{\cos x + 1}\right)\,dx$

$\displaystyle =\int\frac{\cos x + 1}{\cos^2x - 1}\,dx$

Check the spoiler if you're still stuck:

Spoiler:
$\displaystyle =\int\frac{\cos x + 1}{-\sin^2x}\,dx$

$\displaystyle =-\int\frac{\cos x}{\sin^2x}\,dx - \int\frac1{\sin^2x}\,dx$

$\displaystyle =-\int\left(\sin x\right)^{-2}\cos x\,dx - \int\csc^2x\,dx$
• Jun 25th 2012, 05:37 PM
Ragnarok
Re: Integral with trig function in denominator
Thank you! I actually got that far but stopped because I thought I wasn't getting anywhere. How does the rest of the solution go?
• Jun 25th 2012, 05:42 PM
Reckoner
Re: Integral with trig function in denominator
Quote:

Originally Posted by Ragnarok
Thank you! I actually got that far but stopped because I thought I wasn't getting anywhere. How does the rest of the solution go?

For the first integral, let $\displaystyle u = \sin x.$ The second integral should be obvious from the fact that $\displaystyle \textstyle\frac d{dx}\left[\cot x\right] = -\csc^2x.$
• Jun 28th 2012, 02:56 PM
Ragnarok
Re: Integral with trig function in denominator
Thank you! I see it now.