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: 67

Answer

$$\frac{1}{8}\cos 4x - \frac{1}{{20}}\cos 10x + C$$

Work Step by Step

$$\eqalign{ & \int {\sin 3x\cos 7x} dx \cr & \sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr & = \frac{1}{2}\int {\left( {\sin \left( {\left( { - 4} \right)x} \right) + \sin \left( {\left( {10} \right)x} \right)} \right)} dx \cr & = \frac{1}{2}\int {\left( { - \sin 4x + \sin 10x} \right)} dx \cr & {\text{sum rule}} \cr & = - \frac{1}{2}\int {\sin 4x} dx + \frac{1}{2}\int {\sin 10x} dx \cr & {\text{integrating}} \cr & = - \frac{1}{2}\left( { - \frac{1}{4}\cos 4x} \right) + \frac{1}{2}\left( { - \frac{1}{{10}}\cos 10x} \right) + C \cr & {\text{simplify}} \cr & = \frac{1}{8}\cos 4x - \frac{1}{{20}}\cos 10x + C \cr} $$
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