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: 2

Answer

$\ln |\sqrt{9+x^2}+3x|+C$

Work Step by Step

Let us consider $x =\dfrac{1}{3} \tan \theta $ and $dx= \dfrac{1}{3} \sec^2 \theta d \theta $ Now, the given integral becomes: $\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9+9(\dfrac{1}{9} \tan^2 \theta)}} =\int \dfrac{\sec^2 \theta d \theta }{\sqrt { \sec^2 \theta}}$ or, $=\int \sec \theta d \theta $ Now, integrate. we have $\int \sec \theta d \theta=\ln |\sec \theta +\tan \theta|+C$ Plug in $\tan \theta =3x$ Thus, we have $\ln |\sec \theta +\tan \theta|+C=\ln |\sqrt{9+x^2}+3x|+C$
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