Answer
$$\frac{1}{3}\ln \left| {\sin 3x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\cot 3x} dx \cr
& = \int {\frac{{\cos 3x}}{{\sin 3x}}} dx \cr
& {\text{substitute }}u = \sin 3x,{\text{ }}du = 3\cos 3xdx \cr
& = \int {\frac{{\cos 3x}}{{\sin 3x}}} dx = \int {\frac{{1/3}}{u}} du \cr
& = \frac{1}{3}\int {\frac{1}{u}} du \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{3}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sin 3x \cr
& = \frac{1}{3}\ln \left| {\sin 3x} \right| + C \cr} $$
