Answer
$$\int\sqrt{\cot x}\csc^2xdx=-\frac{2}{3}\sqrt{\cot^3x}+C$$
Work Step by Step
$$A=\int\sqrt{\cot x}\csc^2xdx$$
Let $u=\cot x$
Then we have $du=-\csc^2 xdx$. That means $\csc^2dx=-du$
Also, $\sqrt{\cot x}=\sqrt u=u^{1/2}$
Substitute into $A$: $$A=-\int u^{1/2}du$$ $$A=-\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{2\sqrt{u^3}}{3}+C$$ $$A=-\frac{2}{3}\sqrt{\cot^3x}+C$$