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: 42


$\sin^{-1} x+C$

Work Step by Step

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