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: 21

Answer

$$ - \frac{1}{{9x}}\sqrt {{x^2} + 9} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} + 9} }}} \cr & {\text{substitute }}x = 3\tan \theta ,{\text{ }}dx = 3{\sec ^2}\theta d\theta \cr & = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{9{{\tan }^2}\theta \sqrt {9{{\tan }^2}\theta + 9} }}} \cr & {\text{simplify}} \cr & = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{9{{\tan }^2}\theta \left( 3 \right)\sqrt {{{\tan }^2}\theta + 1} }}} = \frac{1}{9}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta \sec \theta }}} \cr & = \frac{1}{9}\int {\frac{{\sec \theta d\theta }}{{{{\tan }^2}\theta }}} = \frac{1}{9}\int {\frac{1}{{\cos \theta }}\left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)d\theta } \cr & = \frac{1}{9}\int {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}d\theta } \cr & = \frac{1}{9}\left( { - \frac{1}{{\sin \theta }}} \right) + C \cr & = - \frac{1}{9}\csc \theta + C \cr & {\text{so}} \cr & = - \frac{1}{{9x}}\sqrt {{x^2} + 9} + C \cr} $$
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