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: 22

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{{dt}}{{{t^2}\sqrt {9 - {t^2}} }}} \cr & {\text{substitute }}t = 3\sin \theta ,{\text{ }}dt = 3\cos \theta d\theta \cr & \int {\frac{{dt}}{{{t^2}\sqrt {9 - {t^2}} }}} = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr & = \int {\frac{{3\cos \theta d\theta }}{{9{{\sin }^2}\theta \sqrt {9 - 9{{\sin }^2}\theta } }}} = \int {\frac{{3\cos \theta d\theta }}{{9{{\sin }^2}\theta \left( 3 \right)\sqrt {1 - {{\sin }^2}\theta } }}} \cr & = \int {\frac{{\cos \theta d\theta }}{{9{{\sin }^2}\theta \left( {\cos \theta } \right)}}} = \frac{1}{9}\int {{{\csc }^2}\theta } d\theta \cr & {\text{integrating}} \cr & = - \frac{1}{9}\cot \theta + C \cr & so \cr & = - \frac{1}{9}\left( {\frac{{\sqrt {9 - {t^2}} }}{t}} \right) + C \cr & = - \frac{{\sqrt {9 - {t^2}} }}{{9t}} + C \cr} $$
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