University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 6


$\ln |2+\sec y|+c$

Work Step by Step

Solve $\int\dfrac{\sec y \tan y dy}{2+\sec y}$ Let $p=2+\sec y$ and $dp=3 \sec^2 t$ This implies $\int\dfrac{\sec y \tan y dy}{2+\sec y}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=2+\sec y$ Thus, $\int\dfrac{\sec y \tan y dy}{2+\sec y}=\ln |2+\sec y|+c$
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