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: 44


$\sqrt{1-(\ln x)^2}+\ln \dfrac{\ln x}{(1+\sqrt{1-(\ln x)^2})}+C$

Work Step by Step

Consider $\ln x=\sin \theta \implies dx/x= \cos \theta d \theta $ Now, the given integral can be written as: $\int\dfrac{\sqrt{1-\sin^2 \theta}(\cos \theta)}{ \sin \theta} d\theta= \int \csc \theta-\sin \theta d\theta$ or, $=\cos \theta-\ln |\csc d \theta+\cot d \theta|+C$ or, $=\sqrt{1-(\ln x)^2}+\ln \dfrac{\ln x}{(1+\sqrt{1-(\ln x)^2})}+C$
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