Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 422: 106

Answer

$$ y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + 2$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }},\,\,\,\,\,\,\,y\left( 0 \right) = 2 \cr & {\text{separate the variables}} \cr & dy = \left( {\frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }}} \right)dx \cr & {\text{integrate both sides}} \cr & y = \int {\left( {\frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr & y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + C\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{use initial condition }}y\left( 0 \right) = 2 \cr & 2 = {\tan ^{ - 1}}0 - 2{\sin ^{ - 1}}0 + C \cr & C = 2 \cr & \cr & {\text{then}}{\text{, substitute }}C = 2{\text{ in }}\left( {\bf{1}} \right) \cr & y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + 2 \cr} $$
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