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: 22


$\frac{1}{3}\cos \frac{1}{x}+C$

Work Step by Step

$\int \frac{\sin(1/x)}{3x^2}dx$ Let $u=\frac{1}{x}=x^{-1}$. Then $du=-x^{-2}dx=-\frac{1}{x^2}dx$, so $\frac{1}{3x^2}dx=-\frac{1}{3}du$ $=\int\sin u*(-\frac{1}{3})du$ $=-\frac{1}{3}\int\sin u du$ $=-\frac{1}{3}(-\cos u+C)$ $=\frac{1}{3}\cos \frac{1}{x}+C$
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