QJEEM22Ma53ed: JEE Main 2022 Mathematics - Class 12 JEEM22Ma53ed Solution: The value of the integral is equal to __________.

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

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

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

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