Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 38

Answer

$$ - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C$$

Work Step by Step

$$\eqalign{ & \int {\left( {\sin 4t - \sin \frac{t}{4}} \right)} dt \cr & {\text{sum Rule}} \cr & = \int {\sin 4t} dt - \int {\sin \frac{t}{4}} dt \cr & {\text{use the formula for indefinite integrals of trigonometric functions}} \cr & \int {\sin ax} dx = - \frac{1}{a}\cos ax + C \cr & = - \frac{1}{4}\cos 4t - \left( { - \frac{1}{{1/4}}\cos \frac{t}{4}} \right) + C \cr & {\text{Simplify}} \cr & = - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dt}}\left( { - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C} \right) \cr & = - \frac{1}{4}\left( { - 4\sin 4t} \right) + 4\left( { - \frac{1}{4}\sin \frac{t}{4}} \right) + 0 \cr & = \sin 4t - \sin \frac{t}{4} \cr} $$
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