$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ {{{\left( {{\pi \over 2} - x} \right)}^3} - {{3{\pi ^2}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$

Using $$\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} } $$

$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ { - {{\left( {{\pi \over 2} - x} \right)}^3} + {{3{\pi ^4}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$

Adding these two equations, we get

$$2I = {{48} \over {{\pi ^4}}}\int_0^\pi {{{{\pi ^3}} \over 2}\,.\,{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$

$$ \Rightarrow I = {{12} \over \pi }\left[ { - {{\tan }^{ - 1}}(\cos x)} \right]_0^\pi = {{12} \over \pi }\,.\,{\pi \over 2} = 6$$