Answer
$$ -\frac{1}{2}\cot^2 x-\ln \sin x+c$$
Work Step by Step
\begin{align*}
\int \cot^3 xdx&= \int \cot x\cot^2 xdx\\
&= \int \cot x(\csc^2 x-1)dx\\
&= \int \cot x\csc^2 xdx-\int \cot x dx\\
&= -\frac{1}{2}\cot^2 x-\ln \sin x+c
\end{align*}
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 19
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