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: 16

Answer

$\displaystyle \frac{1}{3}\tan^{3}\theta+2\tan\theta-\cot\theta+C$

Work Step by Step

$[\tan\theta]'=-\sec^{2}\theta,$ $\sec^{6}\theta=\sec^{4}\theta\sec^{2}\theta$ use identity: $\tan^{2}\theta=\sec^{2}\theta-1$ $\displaystyle \int\frac{\sec^{6}\theta}{\tan^{2}\theta}d\theta=\int\frac{\left(\tan^{2}\theta+1\right)^{2}\sec^{2}\theta}{\tan^{2}\theta}d\theta\quad =\left[\begin{array}{l} u=\tan\theta\\ du=-\sec^{2}\theta d\theta \end{array}\right]$ $=\displaystyle \int\frac{\left(u^{2}+1\right)^{2}}{u^{2}}du$ $=\displaystyle \int\frac{u^{4}+2u^{2}+1}{u^{2}}du$ $=\displaystyle \int\left(u^{2}+2+\frac{1}{u^{2}}\right)du$ $=\displaystyle \frac{u^{3}}{3}+2u-\frac{1}{u}+C$ $=\displaystyle \frac{1}{3}\tan^{3}\theta+2\tan\theta-\cot\theta+C$
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