Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - Exercises - Page 538: 11


$\displaystyle \sqrt{3}-\frac{\pi}{3}$

Work Step by Step

Trigonometric substitution: $\displaystyle \sqrt{x^{2}-a^{2}}\quad x=a\sec\theta,\quad 0\leq\theta\lt \frac{\pi}{2}$ or $\displaystyle \pi\leq\theta\lt \frac{3\pi}{2}$ Identity: $\sec^{2}\theta-1=\tan^{2}\theta$ $\displaystyle \int_{1}^{2}\frac{\sqrt{x^{2}-1}}{x}dx=\left[\begin{array}{ll} x=\sec\theta & \\ dx=\sec\theta\tan\theta d\theta & \\ & \sqrt{x^{2}-1}=\tan\theta\\ x=1\rightarrow\theta=0, & x=2\rightarrow\theta=\pi/3 \end{array}\right]$ $\displaystyle \int_{0}^{\pi/3}\frac{\tan\theta}{\sec\theta}\sec\theta\tan\theta d\theta$ $=\displaystyle \int_{0}^{\pi/3}\tan^{2}\theta d\theta$ ... apply the identity $=\displaystyle \int_{0}^{\pi/3}\left(\sec^{2}\theta-1\right)d\theta$ $=[\tan\theta-\theta]_{0}^{\pi/3}$ $=\displaystyle \sqrt{3}-\frac{\pi}{3}$
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