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.
$\displaystyle \int\frac1{\cos x - 1}\,dx$
$\displaystyle =\int\left(\frac1{\cos x - 1}\cdot\frac{\cos x + 1}{\cos x + 1}\right)\,dx$
$\displaystyle =\int\frac{\cos x + 1}{\cos^2x - 1}\,dx$
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