Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 579: 26

Answer

$$\sqrt {{x^2} - 9} - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sqrt {{x^2} - 9} }}{x}} dx,{\text{ }}x > 3 \cr & {\text{Let }}x = 3\sec \theta ,{\text{ }}dx = 3\sec \theta \tan \theta d\theta \cr & {\text{Substitute}} \cr & \int {\frac{{\sqrt {{x^2} - 9} }}{x}} dx = \int {\frac{{\sqrt {{{\left( {3\sec \theta } \right)}^2} - 9} }}{{3\sec \theta }}} \left( {3\sec \theta \tan \theta } \right)d\theta \cr & = \int {\sqrt {9{{\sec }^2}\theta - 9} } \left( {\tan \theta } \right)d\theta \cr & = \int {\sqrt {9\left( {{{\sec }^2}\theta - 1} \right)} } \left( {\tan \theta } \right)d\theta \cr & {\text{Use the pythagorean identity se}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta \cr & = \int {\sqrt {9{{\tan }^2}\theta } } \left( {\tan \theta } \right)d\theta \cr & = \int {3\tan \theta } \left( {\tan \theta } \right)d\theta \cr & = \int {3{{\tan }^2}\theta } d\theta \cr & = 3\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr & {\text{Integrating}} \cr & {\text{ = 3}}\left( {\tan \theta - \theta } \right) + C \cr & {\text{ = }}3\tan \theta - 3\theta + C \cr & {\text{We know that }}x = 3\sec \theta ,{\text{ }}\theta = {\sec ^{ - 1}}\left( {\frac{x}{3}} \right),{\text{ tan}}\theta = \frac{{\sqrt {{x^2} - 9} }}{3} \cr & {\text{ = }}3\left( {\frac{{\sqrt {{x^2} - 9} }}{3}} \right) - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr & {\text{ = }}\sqrt {{x^2} - 9} - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr} $$
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