Answer
$$\int^{3\pi/2}_\pi\cot^5\Big(\frac{\theta}{6}\Big)\sec^2\Big(\frac{\theta}{6}\Big) d\theta=12$$
Work Step by Step
$$A=\int^{3\pi/2}_\pi\cot^5\Big(\frac{\theta}{6}\Big)\sec^2\Big(\frac{\theta}{6}\Big) d\theta$$ $$A=\int^{3\pi/2}_\pi\frac{\sec^2\frac{\theta}{6}}{\tan^5\frac{\theta}{6}}d\theta$$
We set $u=\tan\frac{\theta}{6}$, which means $$du=\frac{1}{6}\sec^2\frac{\theta}{6}d\theta$$ $$\sec^2\frac{\theta}{6}d\theta=6du$$
For $\theta=3\pi/2$, we have $u=\tan\frac{(3\pi/2)}{6}=\tan\frac{\pi}{4}=1$
For $\theta=\pi$, we have $u=\tan\frac{\pi}{6}=\frac{\sqrt3}{3}$
Therefore, $$A=6\int^1_{\sqrt3/3}\frac{1}{u^5}du=6\int^1_{\sqrt3/3}u^{-5}du$$ $$A=6\times\frac{u^{-4}}{-4}\Big]^1_{\sqrt3/3}=-\frac{3}{2u^4}\Big]^1_{\sqrt3/3}$$ $$A=-\frac{3}{2}\Big(1-\frac{1}{(\sqrt3/3)^4}\Big)$$ $$A=-\frac{3}{2}\Big(1-\frac{1}{\frac{1}{9}}\Big)$$ $$A=-\frac{3}{2}(1-9)=12$$
