Let $I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\left(\tan x\right)}^{\sqrt{2018}}}dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\left(\frac{\sin x}{\cos x}\right)}^{\sqrt{2018}}}dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\cos x\right)}^{\sqrt{2018}}}{{\left(\cos x\right)}^{\sqrt{2018}}+{\left(\sin x\right)}^{\sqrt{2018}}}dx \dots \left(\mathrm{i}\right)$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\cos {\left(\frac{\pi }{2}-x\right)}^{\sqrt{2018}}}{\cos {\left(\frac{\pi }{2}-x\right)}^{\sqrt{2018}}+\sin {\left(\frac{\pi }{2}-x\right)}^{\sqrt{2018}}}dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2018}}}{{\left(\sin x\right)}^{\sqrt{2018}}+{\left(\cos x\right)}^{\sqrt{2018}}}dx ...\left(\mathrm{ii}\right)$  

Adding equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2018}}+{\left(\cos x\right)}^{\sqrt{2018}}}{{\left(\sin x\right)}^{\sqrt{2018}}+{\left(\cos x\right)}^{\sqrt{2018}}}dx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}dx$

$\Rightarrow 2I=\frac{\pi }{2}-0$

$\Rightarrow I=\frac{\pi }{4}$