Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.3 Trigonometric Substitution - 7.3 Exercises - Page 531: 20


$$\displaystyle{\int\frac{x}{\sqrt{1+x^2}}dx=\sqrt{1+x^2} +C}\\$$

Work Step by Step

$\displaystyle{I=\int\frac{x}{\sqrt{1+x^2}}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=\tan\theta & x^2=\tan^2\theta \\ & \\ \frac{dx}{d\theta}=\sec^2\theta & dx=\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int\frac{\tan \theta}{\sqrt{1+\tan^2\theta}}\sec^2\theta\ d\theta}\\ \displaystyle{I=\int\frac{\tan \theta}{\sec\theta}\sec^2\theta\ d\theta}\\ \displaystyle{I=\int\tan\theta\sec\theta\ d\theta}\\ \displaystyle{I=\sec\theta +C}\\ \displaystyle{I=\sqrt{1+x^2} +C}\\ $
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