Answer
$$ - \cot x + \tan x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\csc }^4}x}}{{{{\cot }^2}x}}dx} \cr
& {\text{split off }}{\csc ^4}x \cr
& = \int {\frac{{{{\csc }^2}x}}{{{{\cot }^2}x}}\left( {{{\csc }^2}x} \right)dx} \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{{{\cot }^2}x + 1}}{{{{\cot }^2}x}}\left( {{{\csc }^2}x} \right)dx} \cr
& = \int {\left( {1 + \frac{1}{{{{\cot }^2}x}}} \right)\left( {{{\csc }^2}x} \right)dx} \cr
& u = \cot x,{\text{ }}du = - {\csc ^2}xdx \cr
& = - \int {\left( {1 + \frac{1}{{{u^2}}}} \right)du} \cr
& {\text{evaluate}} \cr
& = - u + \frac{1}{u} + C \cr
& {\text{replace }}u = \cot x \cr
& = - \cot x + \frac{1}{{\cot x}} + C \cr
& = - \cot x + \tan x + C \cr} $$
