## Calculus: Early Transcendentals (2nd Edition)

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