University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 3


$-\frac{1}{4}cos^{4}x +C$

Work Step by Step

Plug in: t= cos x so that $dx= \frac{dt}{-sin x}$ in the given integral. Then we have $\int cos^{3}xsinx dx= \int t^{3}sinx\frac{dt}{-sinx}$ $= -\int t^{3}dt= -\frac{t^{4}}{4}+C$ Undoing substitution, we get $\int cos^{3}xsinx dx=-\frac{1}{4}cos^{4}x +C$
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