The value of the integral

$${{48} \over {{\pi ^4}}}\int\limits_0^\pi {\left( {{{3\pi {x^2}} \over 2} - {x^3}} \right){{\sin x} \over {1 + {{\cos }^2}x}}dx} $$ is equal to __________.

<p>$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ {{{\left( {{\pi \over 2} - x} \right)}^3} - {{3{\pi ^2}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p> <p>Using $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }$</p> <p>$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ { - {{\left( {{\pi \over 2} - x} \right)}^3} + {{3{\pi ^4}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p> <p>Adding these two equations, we get</p> <p>$$2I = {{48} \over {{\pi ^4}}}\int_0^\pi {{{{\pi ^3}} \over 2}\,.\,{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p> <p>$$ \Rightarrow I = {{12} \over \pi }\left[ { - {{\tan }^{ - 1}}(\cos x)} \right]_0^\pi = {{12} \over \pi }\,.\,{\pi \over 2} = 6$$</p>