Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 513: 27

Answer

$-\displaystyle \frac{1}{2x^{2}}\cos^{-1}(\frac{1}{x^{2}})+\frac{1}{2}\sqrt{1-\frac{1}{x^{4}}}+C$

Work Step by Step

$\displaystyle \int\frac{\cos^{-1}(x^{-2})}{x^{3}}dx=\int\cos^{-1}(x^{-2})x^{-3}dx=\quad \left[\begin{array}{l} u=x^{-2}\\ du=-2x^{-3}dx \end{array}\right]$ $=-\displaystyle \frac{1}{2}\int\cos^{-1}udu$ Table of integrals: $\color{blue}{ 88. \quad \displaystyle \int\cos^{-1}udu=u\cos^{-1}u-\sqrt{1-u^{2}}+C }$ $=-\displaystyle \frac{1}{2}(u\cos^{-1}u-\sqrt{1-u^{2}})+C$ ... bring back x... $=-\displaystyle \frac{1}{2x^{2}}\cos^{-1}(\frac{1}{x^{2}})+\frac{1}{2}\sqrt{1-\frac{1}{x^{4}}}+C$
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