Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 43

Answer

$${t^3} + \frac{1}{2}\tan 2t + C$$

Work Step by Step

$$\eqalign{ & \int {\left( {3{t^2} + {{\sec }^2}2t} \right)} dt \cr & {\text{sum rule}} \cr & = \int {3{t^2}} dt + \int {{{\sec }^2}2t} dt \cr & {\text{use the power rule and}} \cr & {\text{use the formula for indefinite integrals of trigonometric functions}} \cr & = 3\left( {\frac{{{t^3}}}{3}} \right) + \frac{1}{2}\tan 2t + C \cr & {\text{simplify}} \cr & = {t^3} + \frac{1}{2}\tan 2t + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dt}}\left( {{t^3} + \frac{1}{2}\tan 2t + C} \right) \cr & {\text{ = }}\frac{d}{{dt}}\left( {{t^3}} \right) + \frac{d}{{dt}}\left( {\frac{1}{2}\tan 2t} \right) + \frac{d}{{dt}}\left( C \right) \cr & = 3{t^2} + \frac{1}{2}\left( {{{\sec }^2}2t} \right)\left( 2 \right) + 0 \cr & = 3{t^2} + {\sec ^2}2t \cr} $$
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