$\int_0^{\frac{\pi}{2}} \frac{\mathrm{~d} x}{1+(\cot x)^{101}}=$

Consider the integral $I = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+(\cot x)^{101}}$.

Applying the substitution $x \rightarrow \frac{\pi}{2} - x$ yields $I = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+(\tan x)^{101}}$.

Adding these two expressions gives $2I = \int_0^{\frac{\pi}{2}} \left( \frac{1}{1+(\cot x)^{101}} + \frac{1}{1+(\tan x)^{101}} \right) \mathrm{d}x$.

The integrand simplifies to $1$ since $\frac{1}{1+(\cot x)^{101}} + \frac{1}{1+(\tan x)^{101}} = \frac{(\tan x)^{101}}{(\tan x)^{101}+1} + \frac{1}{1+(\tan x)^{101}} = 1$.

Therefore, $2I = \int_0^{\frac{\pi}{2}} 1 \mathrm{d}x = \frac{\pi}{2}$, and $I = \frac{\pi}{4}$.

Final answer: $I = \frac{\pi}{4}$