${\int }_0^{\mathrm{\pi }/2} \frac{1}{1+{\tan }^3\mathrm{x}}\mathrm{dx}=$

$$\begin{gathered} \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{1}{1+{\tan }^3\mathrm{x}}\mathrm{dx} \to \left(1\right) \\ By applying king\text{'}s rule \\ {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx \\ \Rightarrow \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{1}{1+{\tan }^3\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{x}\right)}\mathrm{dx} \\ \Rightarrow \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{1}{1+co{t}^3\mathrm{x}}\mathrm{dx} \\ \Rightarrow I={\int }_0^{\mathrm{\pi }/2} \frac{{\tan }^3\mathrm{x}}{1+{\tan }^3\mathrm{x}}\mathrm{dx}\to \left(2\right) \\ Now by adding \left(1\right) and \left(2\right) \\ \Rightarrow I+I={\int }_0^{\mathrm{\pi }/2} \frac{1+{\tan }^3\mathrm{x}}{1+{\tan }^3\mathrm{x}}\mathrm{dx} \\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1 dx \\ \Rightarrow 2I={\left(x\right)}_0^{\frac{\pi }{2}} \Rightarrow 2I=\frac{\pi }{2} \\ \Rightarrow \mathrm{I}=\mathrm{\pi }/4 \end{gathered}$$