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: 101


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

Work Step by Step

Let $ u=\ln(\sin x)$, then $ du= \frac{\cos x}{\sin x} dx $ and hence $$\int \cot x \ln(\sin x)dx = \int \frac{\cos x \ln(\sin x)}{\sin x}dx=\int udu\\ =\frac{1}{2} u^2+c =\frac{1}{2} ( \ln(\sin x))^2+c.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.