Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 24

Answer

$\frac{1}{3}tan^3x+C$

Work Step by Step

Let u=tanx$=>du=sec^2x$dx =$\int tan^2x sec^2xdx$ =$\int u^2du$ =$\frac{1}{3}u^3+C$ =$\frac{1}{3}tan^3x+C$
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