Answer
$$ - \frac{{{{\csc }^5}x}}{5} + \frac{{{{\csc }^3}x}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^3}x{{\csc }^3}x} dx \cr
& {\text{Write the integrand as}} \cr
& \int {{{\csc }^2}x{{\cot }^2}x\cot x\csc x} dx \cr
& {\text{Use the trigonometric identity }}{\cot ^2}\theta = {\csc ^2}\theta - 1 \cr
& = \int {{{\csc }^2}x\left( {{{\csc }^2}x - 1} \right)} \cot x\csc xdx \cr
& {\text{Multiply}} \cr
& = \int {\left( {{{\csc }^4}x - {{\csc }^2}x} \right)} \cot x\csc xdx \cr
& = \int {\left( {{{\csc }^4}x - {{\csc }^2}x} \right)} \cot x\csc xdx \cr
& = \int {{{\csc }^4}x} \cot x\csc xdx - \int {{{\csc }^2}x} \cot x\csc xdx \cr
& = - \int {{{\csc }^4}x} \left( { - \cot x\csc x} \right)dx + \int {{{\csc }^2}x} \left( { - \cot x\csc x} \right)dx \cr
& {\text{Integrate using the power rule}} \cr
& - \frac{{{{\csc }^5}x}}{5} + \frac{{{{\csc }^3}x}}{3} + C \cr} $$
