University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 35



Work Step by Step

$$A=\int\frac{1}{\theta^2}\sin\frac{1}{\theta}\cos\frac{1}{\theta}d\theta$$ We set $u=\frac{1}{\theta}$ Then $$du=-\frac{1}{\theta^2}d\theta$$ That means $$\frac{1}{\theta^2}d\theta=-du$$ Therefore, $$A=-\int\sin u\cos udu=-\frac{1}{2}\int2\sin u\cos udu$$ Apply the identity $2\sin u\cos u=\sin2u$ $$A=-\frac{1}{2}\int\sin2udu$$ $$A=-\frac{1}{2}\times\Big(-\frac{1}{2}\cos2u\Big)+C=\frac{1}{4}\cos2u+C$$ $$A=\frac{1}{4}\cos\frac{2}{\theta}+C$$
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