Answer
$$\int^{\pi/2}_{\pi/4}\cot tdt=\ln\sqrt2$$
Work Step by Step
$$A=\int^{\pi/2}_{\pi/4}\cot tdt=\int^{\pi/2}_{\pi/4}\frac{\cos t}{\sin t}dt$$
We set $u=\sin t$, which means $$du=\cos tdt$$
For $t=\pi/2$, we have $$u=\sin\frac{\pi}{2}=1$$
For $t=\pi/4$, we have $$u=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$$
Therefore, $$A=\int_{\sqrt2/2}^1\frac{1}{u}du$$ $$A=\ln|u|\Big]_{\sqrt2/2}^1$$ $$A=\ln1-\ln\frac{\sqrt2}{2}=-\ln\frac{\sqrt2}{2}$$ $$A=\ln\Big(\frac{\sqrt2}{2}\Big)^{-1}$$ $$A=\ln\frac{2}{\sqrt2}=\ln\sqrt2$$
