Answer
$$\int^{\pi/3}_{\pi/6}\cot^3xdx=\frac{4}{3}-\ln\sqrt3$$
Work Step by Step
$$A=\int^{\pi/3}_{\pi/6}\cot^3xdx=\int^{\pi/3}_{\pi/6}\cot x\cot^2xdx$$ $$A=\int^{\pi/3}_{\pi/6}\cot x(\csc^2x-1)dx$$ $$A=\int^{\pi/3}_{\pi/6}\cot x\csc^2xdx-\int^{\pi/3}_{\pi/6}\cot xdx$$ $$A=-\int^{\pi/3}_{\pi/6}\cot x d(\cot x)-\ln|\sin x|\Big]^{\pi/3}_{\pi/6}$$ $$A=\int_{\pi/3}^{\pi/6}\cot xd(\cot x)-(\ln|\sin(\pi/3)|-\ln|\sin(\pi/6)|)$$ $$A=\frac{1}{2}(\cot^2x)\Big]_{\pi/3}^{\pi/6}-(\ln(\sqrt3/2)-\ln(1/2))$$ $$A=\frac{1}{2}\Big[(\sqrt3)^2-\Big(\frac{1}{\sqrt3}\Big)^2\Big]-\ln\frac{\sqrt3/2}{1/2}$$ $$A=\frac{1}{2}(3-\frac{1}{3})-\ln\sqrt3$$ $$A=\frac{1}{2}(\frac{8}{3})-\ln\sqrt3$$ $$A=\frac{4}{3}-\ln\sqrt3$$
