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