Answer
$$\frac{1}{2}(\ln (\sin x))^2+c.$$
Work Step by Step
Let $u=\ln (\sin x)$, then $du=\frac{\cos}{\sin x}dx=\cot x \ dx$ and hence we get
$$\int \cot x \ln(\sin x)dx= \int udu=\frac{1}{2}u^2+c \\
=\frac{1}{2}(\ln (\sin x))^2+c.$$
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: 107
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