Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 23


$\frac{1}{5}\cos \frac{5}{x}+C$

Work Step by Step

$\int \frac{\sin(5/x)}{x^2}dx$ Let $u=\frac{5}{x}=5x^{-1}$. Then $du=-5x^{-2}dx=-\frac{5}{x^2}dx$, so $\frac{1}{x^2}dx=-\frac{1}{5}du$. $=\int\sin u*(-\frac{1}{5})du$ $=-\frac{1}{5}\int\sin u du$ $=-\frac{1}{5}(-\cos u+C)$ $=\frac{1}{5}\cos \frac{5}{x}+C$
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