$\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{\mathrm{d} x}{1+\tan ^3 x}=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} \mathrm{~d} x=\frac{\frac{\pi}{2}-0}{2}=\frac{\pi}{4} \\ & {\left[\because \int_a^b \frac{f(x) \mathrm{d} x}{f(x)+f(a+b-x)}=\frac{b-a}{2}\right]}\end{aligned}$