Integral with trig function in denominator

**Ragnarok** · June 25th 2012, 04:19 PM

Another seemingly simple integral I am going blank on:

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

I really appreciate any help! I have tried different things but none of them seem to work.

**Reckoner** · June 25th 2012, 04:34 PM

$\int\frac1{\cos x - 1}\,dx$

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

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

Check the spoiler if you're still stuck:

Spoiler:

**Ragnarok** · June 25th 2012, 05:37 PM

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?

**Reckoner** · June 25th 2012, 05:42 PM

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 $u = \sin x.$ The second integral should be obvious from the fact that $\textstyle\frac d{dx}\left[\cot x\right] = -\csc^2x.$

**Ragnarok** · June 28th 2012, 02:56 PM

Thank you! I see it now.

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