Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

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



Work Step by Step

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