Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 69

Answer

$${\sec ^{ - 1}}\left| {2y} \right| + C $$

Work Step by Step

$$\eqalign{ & \int {\frac{{dy}}{{y\sqrt {4{y^2} - 1} }}} \cr & = \int {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} \cr & {\text{use the substitution method}}{\text{:}} \cr & u = 2y,{\text{ so that }}du = 2dy \cr & {\text{then}} \cr & \int {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} = \int {\frac{{du/2}}{{2y\sqrt {{u^2} - 1} }}} \cr & = \int {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \cr & {\text{intgrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 1 \cr & = {\sec ^{ - 1}}\left| u \right| + C \cr & {\text{replace }}2y{\text{ for }}u \cr & = {\sec ^{ - 1}}\left| {2y} \right| + C \cr} $$
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