Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 47

$$\int \cot ^{3} 2 x d x =\frac{-1}{4}\csc 2x -\frac{1}{2}\ln |\sin 2x|+c$$

$$ \int \cot ^{3} 2 x d x $$ Since $$1+\cot^2x=\csc^2x$$ Then \begin{align*} \int \cot ^{3} 2 x d x&=\int\cot (2x) (\csc ^{2} 2 x-1) d x\\ &=\int[\cot (2x) \csc ^{2} (2 x)-\cot (2x)] d x\\ &=\frac{-1}{4}\csc 2x -\frac{1}{2}\ln |\sin 2x|+c \end{align*}