Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 10

Answer

$$ \frac{-1}{x}\sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) +c$$

Work Step by Step

Given $$\int\frac{\cos(1/x)}{x^3}dx$$ let $z=\dfrac{1}{x}\ \ \to \ dz=\dfrac{-1}{x^2}dx$, then $$\int\frac{\cos(1/x)}{x^3}dx = -\int z\cos zdz $$ Let \begin{align*} u&=z\ \ \ \ \ \ \ \ \ \ \ \ dv=\cos z dz\\ du&=dz\ \ \ \ \ \ \ \ \ \ \ \ v=\sin z \end{align*} Then \begin{align*} \int\frac{\cos(1/x)}{x^3}dx &=- \int z\cos zdz \\ &=-z\sin z +\int \sin z dz\\ &=-z\sin z - \cos z +c\\ &= \frac{-1}{x}\sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) +c \end{align*}
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