Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 27

Answer

$$ - \frac{1}{5}{\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)^5} + C $$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{{{x^6}}}} dx \cr & {\text{Use trigonometric substitutions}} \cr & {\text{ }}x = \sin \theta,{\text{ }}dx = \cos \theta d\theta \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{{{x^6}}}} dx = \int {\frac{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}{{{{\sin }^6}\theta }}} \left( {\cos \theta d\theta } \right) \cr & {\text{Use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr & = \int {\frac{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}\cos \theta }}{{{{\sin }^6}\theta }}} d\theta \cr & {\text{simplifying, we get:}} \cr & = \int {\frac{{{{\cos }^4}\theta }}{{{{\sin }^6}\theta }}} d\theta \cr & = \int {\frac{{{{\cos }^4}\theta }}{{{{\sin }^4}\theta }}} \left( {\frac{1}{{{{\sin }^2}\theta }}} \right)d\theta \cr & = \int {{{\cot }^4}\theta } {\csc ^2}\theta d\theta \cr & {\text{integrate}} \cr & = - \frac{1}{5}{\cot ^5}\theta + C \cr & {\text{write in terms of }}x,{\text{ }}\cot \theta = \frac{{\sqrt {1 - {x^2}} }}{x} \cr & = - \frac{1}{5}{\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)^5} + C \cr} $$
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