Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises: 17

Answer

$\frac{1}{2}\sin^4 x$

Work Step by Step

$\int \sin^2 x\sin 2x\ dx$ $=\int \sin^2 x*2\sin x\cos x\ dx$ $=2\int\sin^3 x\cos x\ dx$ Let $u=\sin x$. Then $du=\cos x\ dx$. $=2\int u^3\ du$ $=2*\frac{u^4}{4}$ $=\boxed{\frac{1}{2}\sin^4 x}$
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