Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 503: 61

Answer

$$ - \frac{1}{3}\ln \left( 5 \right)$$

Work Step by Step

$$\eqalign{ & \int_{ - 2}^2 {\frac{{dt}}{{{t^2} - 9}}} = - \int_{ - 2}^2 {\frac{{dt}}{{9 - {t^2}}}} \cr & {\text{using the Integral formulas in the Theorem 6}}{\text{.12}}:{\text{ }}\int {\frac{{dx}}{{{a^2} - {x^2}}} = \left\{ {\frac{1}{a}{{\tanh }^{ - 1}}\frac{x}{a} + C,{\text{ for }}\left| x \right| < a} \right.} \cr & or{\text{ }}\frac{{dx}}{{{a^2} - {x^2}}} = \left\{ {\frac{1}{a}{{\coth }^{ - 1}}\frac{x}{a} + C,{\text{ for }}\left| x \right| > a} \right. \cr & {\text{Comparing the integral}} - \int_{ - 2}^2 {\frac{{dt}}{{9 - {t^2}}}} {\text{ with }}\int {\frac{{dx}}{{{a^2} - {x^2}}}} {\text{ we can note that }}a = 3{\text{ and the limits of}} \cr & {\text{ the integration are 2}} \leqslant x \leqslant 2{\text{ then }}\left| x \right| < 3.{\text{ So }}\int {\frac{{dx}}{{{a^2} - {x^2}}}} = \frac{1}{a}{\tanh ^{ - 1}}\frac{x}{a} + C \cr & {\text{then}} \cr & - \int_{ - 2}^2 {\frac{{dt}}{{9 - {t^2}}}} = - \left( {\frac{1}{3}{{\tanh }^{ - 1}}\frac{x}{3}} \right)_{ - 2}^2 \cr & or \cr & = \left( {\frac{1}{3}{{\tanh }^{ - 1}}\frac{x}{3}} \right)_2^{ - 2} \cr & {\text{evaluate the limits of integration}} \cr & = \frac{1}{3}{\tanh ^{ - 1}}\left( { - \frac{2}{3}} \right) - \frac{1}{3}{\tanh ^{ - 1}}\left( {\frac{2}{3}} \right) \cr & {\text{where tan}}{{\text{h}}^{ - 1}}\left( { - x} \right) = - {\tanh ^{ - 1}}x \cr & = - \frac{1}{3}{\tanh ^{ - 1}}\left( {\frac{2}{3}} \right) - \frac{1}{3}{\tanh ^{ - 1}}\left( {\frac{2}{3}} \right) \cr & = - \frac{2}{3}{\tanh ^{ - 1}}\left( {\frac{2}{3}} \right) \cr & {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}:{\text{ }} \cr & {\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right){\text{ }}\left( {\left| x \right| < 1} \right){\text{ then}}{\text{,}} \cr & = - \frac{2}{3}\left[ {\frac{1}{2}\ln \left( {\frac{{1 + 2/3}}{{1 - 2/3}}} \right)} \right] \cr & = - \frac{1}{3}\ln \left( {\frac{{5/3}}{{1/3}}} \right) \cr & = - \frac{1}{3}\ln \left( 5 \right) \cr} $$
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