Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises: 30

Answer

$\sec{y}-\tan{y}+C$

Work Step by Step

$\int\sec{y}(\tan{y}-\sec{y})dy=\int\sec{y}\tan{y}dy-\int\sec^2{y}dy$ $=\sec{y}-\tan{y}+C$ CHECK: $\frac{d}{dy}(\sec{y}-\tan{y}+C)=\frac{d}{dy}\sec{y}-\frac{d}{dy}\tan{y}+\frac{d}{dy}C$ $=\sec{y}\tan{y}-\sec^2{y}+0=\sec{y}\tan{y}-\sec^2{y}=\sec{y}(\tan{y}-\sec{y})$
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