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: 36



Work Step by Step

$$A=\int\frac{\cos\sqrt\theta}{\sqrt\theta\sin^2\sqrt\theta}d\theta$$ We set $u=\sin\sqrt\theta$ Then $$du=(\sqrt\theta)'\cos\sqrt\theta d\theta=\frac{\cos\sqrt\theta}{2\sqrt\theta}d\theta$$ That means $$\frac{\cos\sqrt\theta}{\sqrt\theta}d\theta=2du$$ Therefore, $$A=2\int\frac{1}{u^2}du=-2\int-\frac{1}{u^2}du$$ $$A=-2\times\frac{1}{u}+C=-\frac{2}{u}+C$$ $$A=-\frac{2}{\sin\sqrt{\theta}}+C$$
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