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: 74

Answer

$a.\quad\sinh^{-1}1$ $b.\quad\ln(1+\sqrt{2})$

Work Step by Step

Substitute:$\quad \left[\begin{array}{ll} u=\ln xx & du=\frac{1}{x}dx\\ x=1 & \rightarrow u=0\\ x=e & \rightarrow u=1 \end{array}\right]$ $I=\displaystyle \int_{1}^{e}\frac{dx}{x\sqrt{1+(\ln x)^{2}}}=\int_{0}^{1}\frac{du}{\sqrt{1+u^{2}}}$ $(a)$ Using the table "lntegrals leading to inverse hyperbolic functions" $\displaystyle \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C,\qquad (a=1)$ $I=[\sinh^{-1}u]_{0}^{1}$ $=\sinh^{-1}1-\sinh^{-1}0$ $=\sinh^{-1}1-0$ $=\sinh^{-1}1$ $(b)$ Using the formulas in the box above these exercises, by which $\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1}),$ We would have $\displaystyle \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\ln[\frac{u}{a}+\sqrt{(\frac{u}{a})^{2}+1}]+C$ So, with $a=1,$ $I=[\ln(u+\sqrt{(u)^{2}+1}]_{0}^{1}$ $=\ln(1+\sqrt{1^{2}+1})-\ln(0+\sqrt{0^{2}+1})$ $=\ln(1+\sqrt{2})-\ln(1)$ $=\ln(1+\sqrt{2})$
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