Answer
$\frac {sin^5x} 5 -\frac {sin^7x} 7 +C$
Work Step by Step
$\int cosx(1-sin^2x)sin^4xdx$
$u=sinx, du=cosxdx$
$\int (u^4-u^6)du$
$\frac {u^5} 5 -\frac {u^7} 7 +C$
$\frac {sin^5x} 5 -\frac {sin^7x} 7 +C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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: 2
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