Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 94

Answer

$\frac{1}{2}\ln|2\sin x+3|+C$

Work Step by Step

We have $$\int \frac{\cos x}{2\sin x+3}dx =\frac{1}{2} \int \frac{2\cos x}{2\sin x+3}dx\\ =\frac{1}{2}\ln|2\sin x+3|+c.$$ where we used the facts that $(\sin u)'=\cos u $ and $\int \frac{u'}{u}=\ln u $.
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