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: 16

Answer

$$\frac{x}{{108\sqrt {4 - {x^2}} }} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\left( {36 - 9{x^2}} \right)}^{ - 3/2}}dx} \cr & {\text{substitute 3}}x = 6\sin \theta \cr & x = 2\sin \theta ,{\text{ }}dx = 2\cos \theta d\theta \cr & = \int {{{\left( {36 - 9{{\left( {2\sin \theta } \right)}^2}} \right)}^{ - 3/2}}\left( {2\cos \theta d\theta } \right)} \cr & = \int {{{\left( {36 - 9\left( {4{{\sin }^2}\theta } \right)} \right)}^{ - 3/2}}\left( {2\cos \theta } \right)d\theta } \cr & = 2\int {{{\left( {36 - 36{{\sin }^2}\theta } \right)}^{ - 3/2}}\cos \theta d\theta } \cr & = 2\int {{{\left( {{6^2}{{\cos }^2}\theta } \right)}^{ - 3/2}}\cos \theta d\theta } \cr & = 2\int {\left( {{6^{ - 3}}{{\cos }^{ - 3}}\theta } \right)\cos \theta d\theta } \cr & = \frac{2}{{{6^3}}}\int {{{\sec }^2}\theta d\theta } \cr & {\text{integrating}} \cr & = \frac{1}{{108}}\tan \theta + C \cr & {\text{substituting tan}}\theta \cr & = \frac{1}{{108}}\left( {\frac{{3x}}{{\sqrt {36 - 9{x^2}} }}} \right) + C \cr & {\text{simplify}} \cr & = \frac{1}{{108}}\left( {\frac{{3x}}{{3\sqrt {4 - {x^2}} }}} \right) + C \cr & = \frac{x}{{108\sqrt {4 - {x^2}} }} + C \cr} $$
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