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 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 431: 70

Answer

$a.\displaystyle \quad\tanh^{-1}(\frac{1}{2})$ $b.\displaystyle \quad\frac{1}{2}\ln 3$

Work Step by Step

$(a)$ Use table: "Integrals leading to inverse hyperbolic functions" 3. $\displaystyle \int\frac{du}{a^{2}-u^{2}}=\left\{\begin{array}{ll} \frac{1}{a}\tanh^{-1}(\frac{u}{a})+C, & u^{2} \lt a^{2}\\ \frac{1}{a}\coth^{-1}(\frac{u}{a})+C, & u^{2}\gt a^{2} \end{array}\right.$ On the interval $[0,1/2]$ the graph of $y=x^{2}$ is below the graph of $y=x$, so we apply the $\tanh^{-1}$ case. $a=1, u(x)=x, du=dx,$ $\displaystyle \int_{0}^{1}\frac{dx}{1-x^{2}}= \left[ \tanh^{-1}(\frac{x}{1}) \right]_{0}^{1/2}$ $= \displaystyle \tanh^{-1}(\frac{1}{2}) - \tanh^{-1}(0)$ $= \displaystyle \tanh^{-1}(\frac{1}{2})$ $(b)$ Using the formulas in the box above these exercises, $\displaystyle \tanh^{-1}x=\frac{1}{2}\ln \displaystyle \frac{1+x}{1-x},\quad |x| \lt 1$ $\displaystyle \tanh^{-1}(\frac{1}{2})=\frac{1}{2}\ln \displaystyle \frac{1+\frac{1}{2}}{1-\frac{1}{2}}$ $=\displaystyle \frac{1}{2}\ln \displaystyle \frac{\frac{3}{2}}{\frac{1}{2}}$ $=\displaystyle \frac{1}{2}\ln 3$
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