University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 44


$\ln |\sec a+\tan a|+c=\ln |\sec e^t +\tan e^t|+c$

Work Step by Step

Consider the integral $\int e^t \sqrt {\tan^2 e^t+1}dt$ Plug in $a=e^t \implies e^t dt =da$ The given integral can be re-written as: $\int \sqrt {\tan^2 a+1}dt=\int \sqrt {sec^2 a} da$ This implies that $\sec (a) da=\ln |\sec a+\tan a|+c=\ln |\sec e^t +\tan e^t|+c$
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