Answer
$$ - \frac{{{{\csc }^2}x}}{2} - \ln \left| {\sin x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^3}x} dx \cr
& {\text{Write the integrand as}} \cr
& = \int {{{\cot }^2}x} \cot xdx \cr
& {\text{Use the trigonometric identity }}{\cot ^2}\theta = {\csc ^2}\theta - 1 \cr
& = \int {\left( {{{\csc }^2}x - 1} \right)} \cot xdx \cr
& {\text{Multiply}} \cr
& = \int {\left( {{{\csc }^2}x\cot x - \cot x} \right)} dx \cr
& = \int {{{\csc }^2}x\cot x} dx - \int {\cot x} dx \cr
& = - \int {\csc x\left( { - \csc x\cot x} \right)} dx - \int {\frac{{\cos x}}{{\sin x}}} dx \cr
& {\text{Integrate}} \cr
& = - \frac{{{{\csc }^2}x}}{2} - \ln \left| {\sin x} \right| + C \cr} $$
