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: 5


$\ln |6+3 \tan t|+c$

Work Step by Step

Solve $\int\dfrac{3 \sec^2 t}{6+3 \tan t}$ Let $p=6+3 \tan t$ and $dp=3 \sec^2 t$ This implies $\int\dfrac{3 \sec^2 t}{6+3 \tan t}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=6+3 \tan t$ Thus, $\int\dfrac{3 \sec^2 t}{6+3 \tan t}=\ln |6+3 \tan t|+c$
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