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: 8

Answer

$$\displaystyle{\int\frac{1}{t^2\sqrt{t^2-16}}dt=\frac{\sqrt{t^2-16}}{ 16t}+C}$$

Work Step by Step

$\displaystyle{I=\int\frac{1}{t^2\sqrt{t^2-16}}dt}\\$ $\displaystyle \left[\begin{array}{ll} t=4\sec\theta & t^2=16\sec^2\theta \\ & \\ \frac{dt}{d\theta}=4\sec\theta\tan\theta & dt=4\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int\frac{1}{16\sec^2\theta\sqrt{16\sec^2\theta-16}}4\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{1}{4\sec\theta\sqrt{16\left(\sec^2\theta-1\right)}}\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{1}{16\sec\theta}\ d\theta}\\ \displaystyle{I=\frac{1}{16}\int\cos\theta\ d\theta}\\ \displaystyle{I=\frac{1}{16}\sin\theta+C}\\$ $\sin\theta=\frac{\sqrt{t^2-16}}{ t}\\$ $\displaystyle{I=\frac{\sqrt{t^2-16}}{ 16t}+C}\\$
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