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: 41

Answer

$$\frac{x}{{\sqrt {100 - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{x}{{10}}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{x^2}}}{{{{\left( {100 - {x^2}} \right)}^{3/2}}}}dx} \cr & {\text{substitute }}x = 10\sin \theta ,{\text{ }}dx = 10\cos \theta d\theta \cr & \int {\frac{{{x^2}}}{{{{\left( {100 - {x^2}} \right)}^{3/2}}}}dx} = \int {\frac{{100{{\sin }^2}\theta }}{{{{\left( {100 - 100{{\sin }^2}\theta } \right)}^{3/2}}}}\left( {10\cos \theta } \right)d\theta } \cr & {\text{simplify}} \cr & = \int {\frac{{100{{\sin }^2}\theta }}{{1000{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}\left( {10\cos \theta } \right)d\theta } \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{100{{\sin }^2}\theta }}{{1000{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}\left( {10\cos \theta } \right)d\theta } \cr & = \int {\frac{{{{\sin }^2}\theta }}{{{{\cos }^3}\theta }}\left( {\cos \theta } \right)d\theta } \cr & = \int {{{\tan }^2}\theta d\theta } \cr & = \int {\left( {{{\sec }^2}\theta - 1} \right)d\theta } \cr & {\text{integrating}} \cr & = \tan \theta - \theta + C \cr & = \frac{x}{{\sqrt {100 - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{x}{{10}}} \right) + C \cr} $$
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