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: 43

Answer

$$\frac{{81}}{{2\left( {81 - {x^2}} \right)}} + \ln \sqrt {81 - {x^2}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{x^3}}}{{{{\left( {81 - {x^2}} \right)}^2}}}dx} \cr & {\text{substitute }}x = 9\sin \theta ,{\text{ }}dx = 9\cos \theta d\theta \cr & \int {\frac{{{x^3}}}{{{{\left( {81 - {x^2}} \right)}^2}}}dx} = \int {\frac{{729{{\sin }^3}\theta }}{{{{\left( {81 - 81{{\sin }^2}\theta } \right)}^2}}}\left( {9\cos \theta d\theta } \right)} \cr & {\text{simplify}} \cr & = \int {\frac{{729{{\sin }^3}\theta }}{{6561{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}\left( {9\cos \theta d\theta } \right)} \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{{{\sin }^3}\theta \cos \theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^2}}}d\theta } \cr & = \int {\frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}d\theta } = \int {{{\tan }^3}\theta d\theta } \cr & = \int {{{\tan }^2}\theta \tan \theta d\theta } = \int {\left( {{{\sec }^2}\theta - 1} \right)\tan \theta d\theta } \cr & = \int {\left( {{{\sec }^2}\theta \tan \theta - \tan \theta } \right)d\theta } \cr & = \int {\left( {\sec \theta \sec \theta \tan \theta - \tan \theta } \right)d\theta } \cr & {\text{sum rule}} \cr & = \int {\sec \theta \sec \theta \tan \theta d\theta } - \int {\tan \theta d\theta } \cr & {\text{integrating}} \cr & = \frac{{{{\sec }^2}\theta }}{2} + \ln \left| {\cos \theta } \right| + C \cr & = \frac{1}{2}{\left( {\frac{9}{{\sqrt {81 - {x^2}} }}} \right)^2} + \ln \left| {\frac{{\sqrt {81 - {x^2}} }}{9}} \right| + C \cr & = \frac{{81}}{{2\left( {81 - {x^2}} \right)}} + \ln \left| {\frac{{\sqrt {81 - {x^2}} }}{9}} \right| + C \cr & {\text{by log properties}} \cr & = \frac{{81}}{{2\left( {81 - {x^2}} \right)}} + \ln \sqrt {81 - {x^2}} + C \cr} $$
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