Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 524: 14


$$\frac{1}{2}\left(\frac{1}{2}\sin \frac{2}{t}-\frac{1}{t} \right)+c$$

Work Step by Step

Given $$ \int \frac{\sin^2(1/t)}{t^2}dt $$ Let $x=\dfrac{1}{t}\ \to\ dx=\dfrac{-dt}{t^2}$, then \begin{align*} \int \frac{\sin^2(1/t)}{t^2}dt &=-\int \sin^2 xdx\\ &=\frac{-1}{2}\int (1-\cos 2x)dx\\ &=\frac{-1}{2}\left(x-\frac{1}{2}\sin 2x\right)+c\\ &=\frac{1}{2}\left(\frac{1}{2}\sin \frac{2}{t}-\frac{1}{t} \right)+c \end{align*}
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