Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 15

a) $\frac{-1}{4}cot^2 2\theta+C$ B) $\frac{-1}{4}csc^2 2\theta+C$

(a)Let $u=cot2\theta=>du=-2-csc^2 2\theta d \theta$ $=>\frac{-1}{2}du=csc^2 2\theta d\theta$ $\int csc^2 2\theta \cot 2\theta d \theta=-\int \frac{1}{2}udu$ $\frac{-1}{2}(\frac{u^2}{4})+C$ $\frac{-u^2}{4}+C$ =$\frac{1}{4}cot^2 2\theta+C$ --- b)Let $u=csc2\theta=>du=-2-csc^2 2\theta cot 2\theta d \theta$ $=>\frac{-1}{2}du=csc^2 2\theta cot 2\theta d\theta$ $\int csc^2 2\theta \cot 2\theta d \theta=-\int \frac{1}{2}udu$ $\frac{-1}{2}(\frac{u^2}{4})+C$ $\frac{-u^2}{4}+C$ =$\frac{1}{4}csc^2 2\theta+C$