Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 5

Answer

$-\frac {cos^3x} 3 + \frac{cos^5x} 5+C$

Work Step by Step

$\int sin^3x cos^2xdx$ $\int (1-cos^2x)cos^2xsinxdx$ $\int (cos^2x-cos^4x)sinxdx$ $-\int (cos^2-cos^4)(-sinx)dx$ $-\frac {cos^3x} 3 + \frac{cos^5x} 5+C$
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