Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 50

$$-\frac{8}{3} \cot ^{3} t+8 \cot t+8 t+C $$

We integrate as follows: \begin{align*} \int 8 \cot ^{4} t d t&=8 \int\left(\csc ^{2} t-1\right) \cot ^{2} t d t\\ &=8 \int \csc ^{2} t \cot ^{2} t d t-8 \int \cot ^{2} t d t\\ &=-\frac{8}{3} \cot ^{3} t-8 \int\left(\csc ^{2} t-1\right) d t\\ &=-\frac{8}{3} \cot ^{3} t+8 \cot t+8 t+C \end{align*}