Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 63

$\frac{1}{2}\cot(9-2\theta)+c.$

Since $(\cot(9-2\theta))'=2\csc^2(9-2\theta)$, we have \begin{align*} \int\csc^2(9-2\theta) d\theta &=\frac{1}{2}\int 2\csc^2(9-2\theta) d\theta \\ &=\frac{1}{2}\cot(9-2\theta)+c. \end{align*}