Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - Chapter 7 Review Exercises - Page 558: 16

Answer

$$\frac{{{{\cos }^5}2x}}{{10}} - \frac{{{{\cos }^3}2x}}{6} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sin }^3}2x{{\cos }^2}2x} dx \cr & {\text{split off }}{\sin ^3}2x \cr & = \int {{{\sin }^2}2x{{\cos }^2}} 2x\sin 2xdx \cr & {\text{identity }}{\sin ^2}2x + {\cos ^2}2x = 1 \cr & = \int {{{\sin }^2}2x{{\cos }^2}2x} \sin 2xdx \cr & = \int {\left( {1 - {{\cos }^2}2x} \right)} co{s^2}2x\sin 2xdx \cr & = \int {\left( {{{\cos }^2}2x - {{\cos }^4}2x} \right)} \sin 2xdx \cr & {\text{substitute }}u = \cos 2x,{\text{ }}du = - 2\sin xdx \cr & = \int {\left( {{u^2} - {u^4}} \right)} \left( { - \frac{1}{2}du} \right) \cr & = \frac{1}{2}\int {\left( {{u^4} - {u^2}} \right)} du \cr & {\text{find the antiderivatives by the power rule}} \cr & = \frac{{{u^5}}}{{10}} - \frac{{{u^3}}}{6} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \cos 2x \cr & = \frac{{{{\cos }^5}2x}}{{10}} - \frac{{{{\cos }^3}2x}}{6} + C \cr} $$
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