Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 104


$$\frac{1}{\ln 5}5^{\sec \theta}+c$$

Work Step by Step

Let $u=\sec \theta$, then $du=\sec \theta \tan \theta d\theta$ and hence we get $$\int (\sec \theta \tan \theta) 5^{\sec \theta }d\theta=\int 5^u du\\ =\frac{1}{\ln 5}5^u+c=\frac{1}{\ln 5}5^{\sec \theta}+c.$$
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