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: 32


$2\sqrt{\sec z}+C $

Work Step by Step

Use the u-substitution: $u=\sec z$ Derive: $du=\sec z \tan z dz $ Subsitute: $\int \frac{\sec z \tan z}{\sqrt{\sec z}}dz $ =$\int \frac{1}{\sqrt{u}}du $ =$\int u^{-1/2}du $ =$2u^{1/2}+C $ =$2\sqrt{\sec z}+C $
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