Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Review - Exercises - Page 430: 29

Answer

$\int sin~\pi t~cos~\pi t~dt = \frac{1}{2\pi}~(sin~\pi t)^2+C$

Work Step by Step

$\int sin~\pi t~cos~\pi t~dt$ Let $u = sin~\pi t$ $\frac{du}{dt} = \pi~cos~\pi t$ $dt = \frac{du}{\pi~cos~\pi t}$ $\int (u)(cos~\pi t)~(\frac{du}{\pi~cos~\pi t})$ $=\int \frac{1}{\pi}~u~du$ $=\frac{1}{2\pi}~u^2+C$ $=\frac{1}{2\pi}~(sin~\pi t)^2+C$
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