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: 69

Answer

$a.\displaystyle \quad\coth^{-1}(2) - \coth^{-1}(\frac{5}{4})$ $b.\displaystyle \quad-\frac{\ln 3}{2}$

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 $[\displaystyle \frac{5}{4},2]$, the graph of $x^{2}$ is above the graph of $x$, so we apply the $\coth^{-1}$ case. $a=1, u(x)=x, du=dx,$ $\displaystyle \int_{5/4}^{2}\frac{dx}{1-x^{2}}= \left[ \coth^{-1}(\frac{x}{1}) \right]_{5/4}^{2}$ $= \displaystyle \coth^{-1}(2) - \coth^{-1}(\frac{5}{4})$ $ (b)$ Using the formulas in the box above these exercises, $\displaystyle \coth^{-1}x=\frac{1}{2}\ln\frac{x+1}{x-1},\quad |x| \gt 1$ $\displaystyle \coth^{-1}(2) - \coth^{-1}(\frac{5}{4})=\frac{1}{2}\ln\frac{2+1}{2-1}-\frac{1}{2}\ln\frac{\frac{5}{4}+1}{\frac{5}{4}-1}$ $=\displaystyle \frac{1}{2}[\ln 3-\ln\frac{\frac{9}{4}}{\frac{1}{4}}]$ $=\displaystyle \frac{1}{2}[\ln 3-\ln 9]\qquad $... ( $9=3^{2}$ ) $=\displaystyle \frac{1}{2}[\ln 3-2\ln 3]$ $=-\displaystyle \frac{\ln 3}{2}$
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