Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 52

$$\int \frac{\cot ^{3} t}{\csc t} d t =-\csc t-\sin t+c$$

$$ \int \frac{\cot ^{3} t}{\csc t} d t $$ Since \begin{align*} \int \frac{\cot ^{3} t}{\csc t} d t&= \int \frac{(\csc ^{2} t-1)\cot t}{\csc t} d t\\ &=\int (\csc t\cot t-\cos t)dt\\ &=-\csc t-\sin t+c \end{align*}