Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 409: 1


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Work Step by Step

(a) Since $ x=3\sin \theta $ then $ dx=3\cos\theta d\theta $ and hence $$ I=\int \frac{dx}{\sqrt{9-x^2}}=\int \frac{3\cos\theta d\theta }{\sqrt{9-9\sin^2\theta}}=\int \frac{3\cos\theta d\theta }{\sqrt{9(1-\sin^2\theta)}}\\ =\int \frac{3\cos\theta d\theta }{\sqrt{9\cos^2\theta}}=\int d\theta.$$ (b) $$ I=\int d\theta +c=\theta+c=\sin^{-1}\frac{x}{3}+c. $$
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