Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 35

$\frac{-1}{2}sin^2\frac{1}{\theta}+C$

Let u=$sin\frac{1}{\theta}=>du=(\cos \frac{1}{\theta})(\frac{-1}{theta})d\theta=>-du=\frac{1}{\theta^2}cos \frac{1}\theta d\theta$ =$\int \frac{1}{\theta^2} \sin \frac{1}{\theta} \cos\frac{1}{\theta}d \theta$ =$\int -u du$ =$\frac{-1}{2}u^2+C$ =$\frac{-1}{2}sin^2\frac{1}{\theta}+C$