Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.3 - Trigonometric Substitution - 7.3 Exercises - Page 505: 8

Answer

$ - \ln \left| {\frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }}} \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sqrt {2 - {x^2}} }}{{{x^2}}}dx} ,{\text{ }}x = \sqrt 2 \sin \theta \cr & {\text{Let }}x = \sqrt 2 \sin \theta ,{\text{ }}dx = x = \sqrt 2 \cos \theta d\theta \cr & {\text{Substituting}} \cr & \int {\frac{{\sqrt {2 - {x^2}} }}{{{x^2}}}dx} = \int {\frac{{\sqrt {2 - {{\left( {\sqrt 2 \cos \theta } \right)}^2}} }}{{{{\left( {\sqrt 2 \cos \theta } \right)}^2}}}\left( {\sqrt 2 \cos \theta } \right)d\theta } \cr & {\text{Simplify the integrand}} \cr & = \int {\frac{{\sqrt {2 - 2{{\cos }^2}\theta } }}{{2{{\cos }^2}\theta }}\left( {\sqrt 2 \cos \theta } \right)d\theta } \cr & = \frac{{\sqrt 2 }}{2}\int {\frac{{\sqrt 2 \sqrt {1 - {{\cos }^2}\theta } }}{{{{\cos }^2}\theta }}\left( {\cos \theta } \right)d\theta } \cr & = \int {\frac{{\sqrt {1 - {{\cos }^2}\theta } }}{{\cos \theta }}d\theta } \cr & {\text{Use the identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr & = \int {\frac{{\sqrt {{{\sin }^2}\theta } }}{{\cos \theta }}d\theta } \cr & = \int {\frac{{\sin \theta }}{{\cos \theta }}d\theta } \cr & {\text{Integrating}} \cr & = - \ln \left| {\cos \theta } \right| + C \cr & {\text{Write in terms of }}x,{\text{ use the triangle shown below}} \cr & \cos \theta = \frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }} \cr & = - \ln \left| {\frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }}} \right| + C \cr} $$
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