University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 24



Work Step by Step

$$A=\int\tan^2x\sec^2xdx$$ We set $u=\tan x$ Then $$du=\sec^2xdx$$ Therefore, $$A=\int u^2du=\frac{u^3}{3}+C$$ $$A=\frac{\tan^3x}{3}+C$$
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