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: 68

Answer

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

Work Step by Step

$(a)$ Use table: "Integrals leading to inverse hyperbolic functions" 1. $\displaystyle \quad \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C, \quad a\gt 0$ Here, a=$1$, $u(x)=3x.\displaystyle \qquad du=3dx, \quad (dx=\frac{du}{3})$ The integral bounds change to $\left\{\begin{array}{lll} x=\frac{1}{3} & \rightarrow & u=1\\ x=0 & \rightarrow & u=0 \end{array}\right\}$ $\displaystyle \int_{0}^{1/3}\frac{6dx}{\sqrt{1+(3x)^{2}}} =\int_{0}^{1}\frac{2\cdot du}{\sqrt{1^{2}+u^{2}}}$ $=2\displaystyle \cdot \left[ \sinh^{-1}(\frac{u}{1}) \right]_{0}^{1}$ $=2[\sinh^{-1}(1)-\sinh^{-1}0]\qquad $... ( $\sinh 0=0 )$ $=2\sinh^{-1}1$ $(b)$ Using the formulas in the box above these exercises, $\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1})$ ,$\quad -\infty \lt x \lt \infty$ $2\sinh^{-1}1=2\ln(1+\sqrt{1+1})$ $=2\ln(1+\sqrt{2})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.