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: 2

Answer

$\frac {sin^5x} 5 -\frac {sin^7x} 7 +C$

Work Step by Step

$\int cosx(1-sin^2x)sin^4xdx$ $u=sinx, du=cosxdx$ $\int (u^4-u^6)du$ $\frac {u^5} 5 -\frac {u^7} 7 +C$ $\frac {sin^5x} 5 -\frac {sin^7x} 7 +C$
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