Answer
$\frac 1 {16}sin^82x+C$
Work Step by Step
$u=sin2x, du=2cos2xdx$
$\int sin^72xcos2xdx$
$\frac 1 2 \int sin^72x(2cos2x)dx$
$\frac 1 2 (\frac {sin^82x} 8)+C$
$\frac 1 {16}sin^82x+C$
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: 3
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