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: 105

Answer

$$ y = {\sec ^{ - 1}}x + \frac{2}{3}\pi $$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{x\sqrt {{x^2} - 1} }},\,\,\,\,x > 1;\,\,\,\,\,\,y\left( 2 \right) = \pi \cr & {\text{separating the variables, we get:}} \cr & dy = \left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right)dx \cr & {\text{integrate both sides}} \cr & y = \int {\left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right)} dx \cr & y = {\sec ^{ - 1}}x + C\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{use initial condition }}y\left( 2 \right) = \pi \cr & \pi = {\sec ^{ - 1}}\left( 2 \right) + C \cr & \pi = \frac{\pi }{3} + C \cr & C = \pi - \frac{\pi }{3} = \frac{2}{3}\pi \cr & \cr & {\text{then}}{\text{, substitute }}C = \frac{2}{3}\pi {\text{ in }}\left( {\bf{1}} \right) \cr & y = {\sec ^{ - 1}}x + \frac{2}{3}\pi \cr} $$
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