Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 356: 37


$f(t)=\tan t+\sec t-2-\sqrt{2}$

Work Step by Step

$f'(t)=\sec^{2}t+\sec t\tan t$ Using the antiderivatives table, $f(t)=\tan t+\sec t+C$ Given that $f(\pi/4)=-1$, we find $C$: $-1=\tan(\pi/4)+\sec(\pi/4)+C$ $-1=1+\sqrt{2}+C$ $C=-2-\sqrt{2}$ $f(t)=\tan t+\sec t-2-\sqrt{2}$
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