Answer
$$\ln \left| {\sin x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\cot x} dx \cr
& {\text{Using the trigonometric identity }}\cot x = \frac{{\cos x}}{{\sin x}} \cr
& \int {\cot x} dx = \int {\frac{{\cos x}}{{\sin x}}} dx \cr
& {\text{Set }}u = \sin x \to du = \cos xdx \cr
& \int {\frac{{\cos x}}{{\sin x}}} dx = \int {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \ln \left| u \right| + C \cr
& {\text{substituting }}u = \sin x \cr
& = \ln \left| {\sin x} \right| + C \cr} $$