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: 9

Answer

$\frac{1}{2}(sec(2t))+c$

Work Step by Step

$u=2t$, therefore $du=2dt$. Let's rewrite the integral! $\int (sec(2t)tan(2t))dt$ We do the substitution! $\int (sec(2t)tan(2t))dt=\int (sec(u)tan(u))\frac{du}{2}=\frac{1}{2}\int (sec(u)tan(u))du=\frac{1}{2}(secu)+c$ We re-do the substitution! $\frac{1}{2}(secu)+c=\frac{1}{2}(sec(2t))+c$
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