${\int }_0^{\pi /2} \frac{1}{1+{\tan }^{2020}\left(x\right)}dx=$

$I={\int }_0^{\pi /2} \frac{1}{1+{\tan }^{2020}x}dx={\int }_0^{\pi /2} \frac{{\cos }^{2020}x}{{\cos }^{2020}x+{\sin }^{2020}x}dx\to \left(1\right)$

$\begin{array}{l}I={\int }_0^{\pi /2} \frac{{\cos }^{2020}(\pi /2-x)}{{\cos }^{2020}(\pi /2-x)+{\sin }^{2020}(\pi /2-x)}dx\\ ={\int }_0^{\pi /2} \frac{{\sin }^{2020}x}{{\sin }^{2020}x+{\cos }^{2020}x}dx\to \left(2\right)\\ \\ \left.\left(1\right)+\left(2\right)\Rightarrow 2I={\int }_0^{\pi /2} \frac{{\cos }^{2020}x+{\sin }^{2020}x}{{\sin }^{2020}x+{\cos }^{2020}x}\right]dx\\ ={\int }_0^{\pi /2} dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\Rightarrow {\int }_0^{\pi /2} \frac{dx}{1+{\tan }^{2020}x}=\frac{\pi }{4}.\end{array}$