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: 32


$$\int\frac{\sec z\tan z}{\sqrt{\sec z}}dz=2\sqrt{\sec z}+C$$

Work Step by Step

$$A=\int\frac{\sec z\tan z}{\sqrt{\sec z}}dz$$ We set $u=\sqrt{\sec z}$ Then $$du=\frac{(\sec z)'}{2\sqrt{\sec z}}dz=\frac{\sec z\tan z}{2\sqrt{\sec z}}dz$$ That means $$\frac{\sec z\tan z}{\sqrt{\sec z}}dz=2du$$ Therefore, $$A=\int2du=2u+C$$ $$A=2\sqrt{\sec z}+C$$
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