Answer
$\frac{1}{2}$$sec(2x)$ + $c$
Work Step by Step
$u$ = $2x$
$du$ = $2$ $dx$
$\frac{1}{2}$$du$ = $dx$
$\int$ $sec(2x)$$tan(2x)$ $dx$ = $\frac{1}{2}$$\int$$sec(u)tan(u)$ $du$
= $\frac{1}{2}$$sec(u)$ + $c$
= $\frac{1}{2}$$sec(2x)$ + $c$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - Review Exercises - Page 314: 72
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