Answer
$12$
Work Step by Step
Applying th.7, let $u=g(\theta)=\tan(\theta/6)$ (a continuous function).
Then, $g'(\displaystyle \theta)=\sec^{2}(\theta/6)\cdot\frac{1}{6}=\frac{1}{6}\sec^{2}(\theta/6)$
Then, $\quad \left[\begin{array}{ll}
u=\tan(\theta/6) & du=g'(\theta) d\theta\\
\cot^{5}(\theta/6)=\frac{1}{u^{5}}=u^{-5} & du=\frac{1}{6}\sec^{2}(\theta/6)d\theta\\
& \sec^{2}(\theta/6)d\theta=6du
\end{array}\right]$,
and, the borders change to $g(\displaystyle \pi)=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ and $\displaystyle \quad g(3\pi/2)=\tan\frac{\pi}{4}=1$
$\displaystyle \int_{\pi}^{3\pi/2}\cot^{5}(\theta/6)\cdot\sec^{2}(\theta/6)=\int_{\sqrt{3}/3}^{1}u^{-5}\cdot(6du) =6\int_{\sqrt{3}/3}^{1}u^{-5}du$
$=6\displaystyle \left[ \frac{u^{-5+1}}{-5+1} \right]_{\sqrt{3}/3}^{1}$
$=6\displaystyle \left[ \frac{u^{-4}}{-4} \right]_{\sqrt{3}/3}^{1}$
$=-\displaystyle \frac{3}{2}\left[ \frac{1}{u^{4}} \right]_{\sqrt{3}/3}^{1}$
$=-\displaystyle \frac{3}{2}(1- \frac{3^{4}}{(\sqrt{3})^{4}})$
$=-\displaystyle \frac{3}{2}(1- \frac{3^{4}}{3^{2}})$
$=-\displaystyle \frac{3}{2}(-8)$
=$12$
