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: 73


$$\frac{5}{4}\sec 2{x^2} + C$$

Work Step by Step

$$\eqalign{ & \int {5x} \sec 2{x^2}\tan 2{x^2}dx \cr & {\text{set }}u = 2{x^2}{\text{ then }}\frac{{du}}{{dx}} = 4x,\,\,\,\,\,\,\,\,\frac{{du}}{{4x}} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {5x} \sec 2{x^2}\tan 2{x^2}dx = \int {5x\sec u\tan u} \left( {\frac{{du}}{{4x}}} \right) \cr & {\text{cancel the common factor }}{x^4} \cr & = \int {5\sec u\tan u} \left( {\frac{{du}}{4}} \right) \cr & = \frac{5}{4}\int {\sec u\tan u} du \cr & {\text{integrate by using the Basic Trigonometric integral }}\int {\sec x\tan x} dx = \sec x + C \cr & = \frac{5}{4}\sec u + C \cr & {\text{write in terms of }}x \cr & = \frac{5}{4}\sec 2{x^2} + C \cr} $$
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