Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 33


$\int tan~x~ln(cos~x)~dx = -\frac{1}{2}[ln(cos~x)]^2+C$

Work Step by Step

$\int tan~x~ln(cos~x)~dx$ Let $u = ln(cos~x)$ $\frac{du}{dx} = -\frac{sin~x}{cos~x}$ $\frac{du}{dx} = -tan~x$ $dx = -\frac{du}{tan~x}$ $\int (tan~x)~(u)~(-\frac{du}{tan~x})$ $=\int -u~du$ $=-\frac{1}{2}u^2+C$ $=-\frac{1}{2}[ln(cos~x)]^2+C$
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