Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 702: 69


$$\frac{1}{5}\tan 5x + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sec }^2}5x} dx \cr & {\text{set }}u = 5x{\text{ then }}\frac{{du}}{{dx}} = 5,\,\,\,\,\,\,\,\,\frac{{du}}{5} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {{{\sec }^2}5x} dx = \int {{{\sec }^2}u} \left( {\frac{{du}}{5}} \right) \cr & = \frac{1}{5}\int {{{\sec }^2}u} du \cr & {\text{integrate by using the Basic Trigonometric integral }}\int {{{\sec }^2}x} dx = \tan x + C \cr & = \frac{1}{5}\tan u + C \cr & {\text{write in terms of }}x \cr & = \frac{1}{5}\tan 5x + C \cr} $$
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