Answer
$$\frac{1}{2} \ln(2\sin x+3)+c$$
Work Step by Step
Using the fact that $\int u'/u=\ln u$, we have
$$\int \frac{\cos x}{2\sin x+3}dx=\frac{1}{2}\int \frac{2\cos x}{2\sin x+3}dx=\frac{1}{2} \ln(2\sin x+3)+c$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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