Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 69

Answer

$$\frac{1}{2}\sin x - \frac{1}{{10}}\sin 5x + C$$

Work Step by Step

$$\eqalign{ & \int {\sin 3x\sin 2x} dx \cr & \sin mx\sin nx = \frac{1}{2}\left( {\cos \left( {\left( {m - n} \right)x} \right) - \cos \left( {\left( {m + n} \right)x} \right)} \right) \cr & = \frac{1}{2}\int {\left( {\cos \left( {\left( 1 \right)x} \right) - \cos \left( {\left( 5 \right)x} \right)} \right)} dx \cr & = \frac{1}{2}\int {\left( {\cos \left( x \right) - \cos \left( {5x} \right)} \right)} dx \cr & {\text{sum rule}} \cr & = \frac{1}{2}\int {\left( {\cos \left( x \right)} \right)} dx - \frac{1}{2}\int {\left( {\cos \left( {5x} \right)} \right)} dx \cr & {\text{integrating}} \cr & = \frac{1}{2}\sin x - \frac{1}{2}\left( {\frac{1}{5}\sin 5x} \right) + C \cr & {\text{simplify}} \cr & = \frac{1}{2}\sin x - \frac{1}{{10}}\sin 5x + C \cr} $$
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