Answer
$$\frac{{{{\csc }^{10}}x}}{{10}} - \frac{{{{\csc }^{12}}x}}{{12}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\csc }^{10}}x{{\cot }^3}xdx} \cr
& {\text{split the integrand}} \cr
& = \int {{{\csc }^9}x{{\cot }^2}x\csc x\cot xdx} \cr
& {\text{pythagorean identity }}{\cot ^2}x = {\csc ^2}x - 1 \cr
& = \int {{{\csc }^9}x\left( {{{\csc }^2}x - 1} \right)\csc x\cot xdx} \cr
& = \int {\left( {{{\csc }^{11}}x - {{\csc }^9}x} \right)\csc x\cot xdx} \cr
& u = \csc x,{\text{ }}du = - \csc x\cot xdx \cr
& = - \int {\left( {{u^{11}} - {u^9}} \right)du} \cr
& {\text{by the power rule}} \cr
& = \frac{{{u^{10}}}}{{10}} - \frac{{{u^{12}}}}{{12}} + C \cr
& {\text{replace }}u = \csc x \cr
& = \frac{{{{\csc }^{10}}x}}{{10}} - \frac{{{{\csc }^{12}}x}}{{12}} + C \cr} $$
