Calculus with Applications (10th Edition)

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

Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 7

Answer

$$ - \tan 3x + C$$

Work Step by Step

$$\eqalign{ & - \int {3{{\sec }^2}3x} dx \cr & {\text{set }}u = 3x{\text{ then }}\frac{{du}}{{dx}} = 3,\,\,\,\,\,\,\,\,\frac{{du}}{3} = dx \cr & {\text{write the integrand in terms of }}u \cr & - \int {3{{\sec }^2}3x} dx = - \int {3{{\sec }^2}u} \left( {\frac{{du}}{3}} \right) \cr & = - \int {{{\sec }^2}u} du \cr & {\text{integrate by using the Basic Trigonometric integral }}\int {{{\sec }^2}x} dx = \tan x + C \cr & = - \tan u + C \cr & {\text{write in terms of }}x,{\text{ replacing }}3x{\text{ for }}u{\text{ gives}} \cr & = - \tan 3x + C \cr} $$
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