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 - Page 251: 31



Work Step by Step

$\int(\tan^2{y}+1)dy=\int\sec^2{y}dy=\tan{y}+C$ CHECK: $\frac{d}{dy}(\tan{y}+C)=\frac{d}{dy}\tan{y}+\frac{d}{dy}C=\sec^2{y}=0=\sec^2{y}=1+\tan^2{y}$
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