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: 70

Answer

$$\frac{1}{2}\sin x - \frac{1}{6}\sin 3x + C$$

Work Step by Step

$$\eqalign{ & \int {\cos x\cos 2x} dx \cr & \cos mx\cos nx = \frac{1}{2}\left( {\cos \left( {\left( {m - n} \right)x} \right) + \cos \left( {\left( {m + n} \right)x} \right)} \right) \cr & = \frac{1}{2}\int {\left( {\cos \left( {\left( { - 1} \right)x} \right) - \cos \left( {\left( 3 \right)x} \right)} \right)} dx \cr & = \frac{1}{2}\int {\left( {\cos \left( x \right) - \cos \left( {3x} \right)} \right)} dx \cr & {\text{sum rule}} \cr & = \frac{1}{2}\int {\left( {\cos \left( x \right)} \right)} dx - \frac{1}{2}\int {\left( {\cos \left( {3x} \right)} \right)} dx \cr & {\text{integrating}} \cr & = \frac{1}{2}\sin x - \frac{1}{2}\left( {\frac{1}{3}\sin 3x} \right) + C \cr & {\text{simplify}} \cr & = \frac{1}{2}\sin x - \frac{1}{6}\sin 3x + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.