The value of $$\int\limits_0^{\frac{\pi }{2}} {\frac{{dx}}{{1 + {{\tan }^3}x}}} $$   is-

Solution :
$$\eqalign{ & {\text{Let}} \cr & I = \int_0^{\frac{\pi }{2}} {\frac{{dx}}{{1 + {{\tan }^3}x}}} \cr & = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx.....(1)} \cr & I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^3}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}dx} \cr & = \int_0^{\frac{\pi }{2}} {\frac{{si{n^3}x}}{{co{s^3}x + {{\sin }^3}x}}dx.....(2)} \cr & {\text{Adding (1) and (2) we get}} \cr & {\text{2}}I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x + {{\sin }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx} \cr & = \int_0^{\frac{\pi }{2}} {1.dx} = \frac{\pi }{2} \cr & \therefore I = \frac{\pi }{4} \cr} $$