Answer
\[ = \frac{4}{3} - \ln \sqrt 3 + \ln 2 - \ln 2 = \frac{4}{3} - \ln \sqrt 3 \]
Work Step by Step
\[\begin{gathered}
\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {{{\cot }^3}d\theta } \hfill \\
\hfill \\
write\,\,\,\,{\cot ^3}\theta \,\,\,as\,\,{\cot ^2}\theta \cot \theta \hfill \\
\hfill \\
\int_{}^{} {{{\cot }^2}\theta \cot \theta d\theta } \hfill \\
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use\,the\,identity\,{\cot ^2}\theta = {\csc ^2}\theta - 1 \hfill \\
\hfill \\
\int_{}^{} {\,\left( {{{\csc }^2}\theta - 1} \right)\cot \theta d\theta } \hfill \\
\hfill \\
multiply\,\, \hfill \\
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= \int_{}^{} {{{\csc }^2}\theta \cot \theta \,d\theta - \int_{}^{} {\cot \theta d\theta } } \hfill \\
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integrate \hfill \\
\hfill \\
= - \frac{{{{\cot }^2}\theta }}{2} - \ln \left| {\sin \theta } \right| + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {{{\cot }^3}\theta d\theta } = \,\,\left[ { - \frac{1}{2}{{\cot }^2}\theta - \ln \left| {\sin \theta } \right|} \right]_{\frac{\pi }{6}}^{\frac{\pi }{3}} \hfill \\
\hfill \\
Evaluate\,\,\,the\,\,limits \hfill \\
\hfill \\
= - \frac{1}{2}{\cot ^2}\,\left( {\frac{\pi }{3}} \right) - \ln \left| {\sin \frac{\pi }{3}} \right| + \frac{1}{2}{\cot ^2}\,\left( {\frac{\pi }{6}} \right) + \ln \left| {\sin \frac{\pi }{6}} \right| \hfill \\
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simplify \hfill \\
\hfill \\
= - \frac{1}{6} - \ln \,\left( {\frac{{\sqrt 3 }}{2}} \right) + \frac{3}{2} + \ln \,\left( {\frac{1}{2}} \right) \hfill \\
\hfill \\
= \frac{4}{3} - \ln \sqrt 3 + \ln 2 - \ln 2 = \frac{4}{3} - \ln \sqrt 3 \hfill \\
\hfill \\
\end{gathered} \]