Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.2 - Integration by Parts - Exercises 8.2 - Page 455: 3


$t^{2}\sin t+2t\cos t-2\sin t+C$

Work Step by Step

We use the formula $\int udv=uv-\int vdu$ with $u=t^{2}$, $dv=\cos t dt$ $\implies du=2t\,dt$, $v=\sin t$ Then $\int t^{2}\cos t\,dt=t^{2}\sin t-\int 2t\sin tdt$ Now, we should evaluate $\int 2t\sin tdt$ With $u=2t$, $dv=\sin t\,dt$ so that $du=2dt$, $v=-\cos t$ $\int 2t\sin tdt=(2t\times -\cos t)-2\int-\cos t\,dt$ $=-2t\cos t+2\sin t$ Therefore, $\int t^{2}\cos t\,dt=t^{2}\sin t+2t\cos t-2\sin t+C$
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