University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 30


$\tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$

Work Step by Step

Let us consider $t = \dfrac{1}{3} \tan \theta $ and $dt= \dfrac{1}{3} \sec^2 \theta d \theta $ Now, the given integral becomes: $\int \dfrac{6( \dfrac{1}{3} sec^2 \theta)}{\sec ^4 \theta} d\theta=2\int cos^2 \theta d \theta $ Now, integrate: $(\theta+\sin \theta \cos \theta) +C= \tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$
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