If $${I_1} = \int\limits_0^\pi {x\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} $$       and $${I_2} = \pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx,} $$       then :

Solution :
$$\eqalign{ & {\text{Consider }}{I_1} = \int\limits_0^\pi {x\,f\left[ {{{\sin }^3}x + {{\cos }^2}x} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\left( {\pi - x} \right)f\left[ {{{\sin }^3}\left( {\pi - x} \right) + {{\cos }^2}\left( {\pi - x} \right)} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\left( {\pi - x} \right)f\left[ {{{\sin }^3}x + {{\cos }^2}x} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx - } \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx - } {I_1} \cr & \Rightarrow 2{I_1} = \int\limits_0^{\frac{\pi }{2}} {\pi \,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow 2{I_1} = 2\pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow {I_1} = \pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & {I_1} = {I_2}\,\,\,\,\,\left( {{\text{By definition of }}{I_2}} \right) \cr} $$