Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 37

Answer

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

Work Step by Step

$\int \frac{cos(x)}{sin(x)+1}dx$ let $u=sin(x)+1$ $dx=\frac{1}{cos(x)}du$ $\int \frac{cos(x)}{sin(x)+1}dx$ $=\int \frac{1}{u}du$ $=ln|u|+C$ $=ln|sin(x)+1|+C$
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