Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 19

Answer

$$\frac{x}{{\sqrt {1 - {x^2}} }} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}} \cr & {\text{substitute }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr & = \int {\frac{{\cos \theta d\theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \cr & {\text{simplify}} \cr & = \int {\frac{{\cos \theta d\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \cr & = \int {\frac{{\cos \theta d\theta }}{{{{\cos }^3}\theta }}} \cr & = \int {{{\sec }^2}\theta } d\theta \cr & {\text{integrating}} \cr & = \tan \theta + C \cr & {\text{with }}x = \sin \theta ,{\text{ and cos}}\theta = \sqrt {1 - {x^2}} \cr & = \frac{x}{{\sqrt {1 - {x^2}} }} + C \cr} $$
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