Answer
$$\int^{\pi/2}_{\pi/4}(1+e^{\cot\theta})\csc^2\theta d\theta=e$$
Work Step by Step
$$A=\int^{\pi/2}_{\pi/4}(1+e^{\cot\theta})\csc^2\theta d\theta$$
We set $u=\cot\theta$, which means $$du=-\csc^2\theta d\theta$$
For $\theta=\pi/4$, we have $$u=\cot(\pi/4)=1$$
For $\theta=\pi/2$, we have $$u=\cot(\pi/2)=0$$
Therefore, $$A=-\int^0_1(1+e^u)du=\int^{1}_0(1+e^u)du$$ $$A=\int^1_01du+\int^1_0e^udu$$ $$A=u\Big]^1_0+e^u\Big]^1_0$$ $$A=1+(e-e^0)$$ $$A=1+e-1=e$$