Calculus 8th Edition

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

Chapter 4 - Integrals - Review - Exercises - Page 350: 25


$$\frac{1}{2\pi}\sin^2(\pi t)+c$$

Work Step by Step

Given $$\int \sin( \pi t)\cos( \pi t) dt$$ Let $x= \pi t\ \Rightarrow \ dx= \pi dt $ then \begin{align*} \int \sin( \pi t)\cos( \pi t) dt&=\frac{1}{ \pi}\int \sin(x)\cos(x) dx\\ &=\frac{1}{2\pi}\sin^2 x+c\\ &=\frac{1}{2\pi}\sin^2(\pi t)+c \end{align*}
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