Answer
$$\int^{\pi/4}_{\pi/6}\frac{\csc^2xdx}{1+\cot^2x}=\frac{\pi}{12}$$
Work Step by Step
$$A=\int^{\pi/4}_{\pi/6}\frac{\csc^2xdx}{1+\cot^2x}$$
We set $u=\cot x$, which means $$du=-\csc^2x dx$$ $$\csc^2xdx=-du$$
For $x=\pi/4$, we have $$u=\cot\frac{\pi}{4}=1$$
For $x=\pi/6$, we have $$u=\cot\frac{\pi}{6}=\sqrt3$$
Therefore, $$A=-\int^1_{\sqrt3}\frac{1}{1+u^2}du=-\tan^{-1}u\Big]^1_{\sqrt3}$$ $$A=-(\tan^{-1}1-\tan^{-1}\sqrt3)=\tan^{-1}\sqrt3-\tan^{-1}1$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$
