Answer
$\csc x + \cot x + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\cos x - 1}}} \cr
& {\text{Multiplty the numerator and denominator by the conjugate of }} \cr
& {\text{the denominator}} \cr
& = \int {\frac{1}{{\cos x - 1}}} \times \frac{{\cos x + 1}}{{\cos x + 1}}dx \cr
& = \int {\frac{{\cos x + 1}}{{{{\cos }^2}x - 1}}} dx \cr
& {\text{Use the pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int {\frac{{\cos x + 1}}{{ - {{\sin }^2}x}}} dx \cr
& = - \int {\left( {\frac{{\cos x}}{{{{\sin }^2}x}} + \frac{1}{{{{\sin }^2}x}}} \right)} dx \cr
& = - \int {\left( {\csc x\cot x + {{\csc }^2}x} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \left( { - \csc x - \cot x} \right) + C \cr
& = \csc x + \cot x + C \cr} $$