Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 507: 1

Answer

$-\ln (1-\sin x)+C$

Work Step by Step

$\int\frac{\cos x}{1-\sin x}dx$ Let $u=1-\sin x$. Then $du=-\cos x dx$, and $\cos xdx=-du$. $=\int\frac{-du}{u}$ $=-\ln u+C$ $=-\ln (1-\sin x)+C$
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