Answer
$ - \cot x - x + C$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^2}x} dx \cr
& {\text{using the pythagorean identity }}{\cot ^2}x + 1 = cs{c^2}x \cr
& {\text{then}}{\text{, }}{\cot ^2}x = {\csc ^2}x - 1 \cr
& \int {{{\cot }^2}x} dx = \int {\left( {{{\csc }^2}x - 1} \right)} dx \cr
& {\text{sum and difference rule for the integrand}} \cr
& \int {{{\csc }^2}x} dx - \int {dx} \cr
& {\text{using the integration basic rules}} \cr
& - \cot x - x + C \cr} $$
