Answer
$$ - \frac{{{{\cot }^3}\theta }}{3} + \theta + \cot \theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^4}\theta } d\theta \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{{\cot }^n}u = - \frac{{{{\cot }^{n - 1}}u}}{{n - 1}} - \int {{{\cot }^{n - 2}}udu} } \cr
& \int {{{\cot }^4}\theta } d\theta = - \frac{{{{\cot }^{4 - 1}}\theta }}{{4 - 1}} - \int {{{\cot }^{4 - 2}}\theta d\theta } \cr
& = - \frac{{{{\cot }^3}\theta }}{3} - \int {{{\cot }^2}\theta d\theta } \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{{\cot }^2}udu} = - u - \cot u + C \cr
& = - \frac{{{{\cot }^3}\theta }}{3} - \left( { - \theta - \cot \theta + C} \right) \cr
& = - \frac{{{{\cot }^3}\theta }}{3} + \theta + \cot \theta + C \cr} $$
