Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 14

Answer

$$ - \frac{1}{{10}}\ln \left| {\cos 10x} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\tan 10x} dx \cr & {\text{identity tan}}\theta = \frac{{\sin \theta }}{{\cos \theta }} \cr & = \int {\frac{{\sin 10x}}{{\cos 10x}}} dx \cr & {\text{substitute }}u = \cos 10x,{\text{ }}du = - 10\sin 10xdx \cr & = \int {\frac{{ - 1/10}}{u}} du \cr & = - \frac{1}{{10}}\int {\frac{1}{u}} du \cr & {\text{find the antiderivative}} \cr & = - \frac{1}{{10}}\ln \left| u \right| + C \cr & {\text{with }}u = \cos 10x \cr & = - \frac{1}{{10}}\ln \left| {\cos 10x} \right| + C \cr} $$
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