Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 102


$$\frac{1}{2} \ln(2\sin x+3)+c$$

Work Step by Step

Using the fact that $\int u'/u=\ln u$, we have $$\int \frac{\cos x}{2\sin x+3}dx=\frac{1}{2}\int \frac{2\cos x}{2\sin x+3}dx=\frac{1}{2} \ln(2\sin x+3)+c$$
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