The value $I={\int }_0^{\pi /2} \frac{1}{1+\cot x}dx$ is 

Let $f\left(x\right)=\frac{1}{1+\cot x}=\frac{\sin x}{\sin x+\cos x},$

$\begin{array}{l}f(\pi /2-x)=\frac{\sin (\pi /2-x)}{\sin (\pi /2-x)+\cos (\pi /2-x)}\\ =\frac{\cos x}{\cos x+\sin x}\\ \text{ Considering }2I={\int }_0^{\pi /2} f\left(x\right)\mathrm{d}x+{\int }_0^{\pi /2} f(\pi /2-x)\mathrm{d}x\\ ={\int }_0^{\pi /2} \frac{\sin x}{\cos x+\sin x}\mathrm{d}x+{\int }_0^{\pi /2} \frac{\cos x}{\cos x+\sin x}\mathrm{d}x\end{array}$

$={\int }_0^{\pi /2} \frac{\sin x+\cos x}{\cos x+\sin x}\mathrm{d}x={\int }_0^{\pi /2} 1\cdot \mathrm{d}x=\frac{\pi }{2}$

So $I = \mathrm{\pi }/4$