STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

$ = \int_0^{\pi /2} {\frac{{\left( {\frac{\pi }{2} - x} \right)\cos x\sin x}}{{{{\sin }^4}x + {{\cos }^4}x}}} $.....$(ii)$

By adding $(i)$ and $(ii),$ we get

$2I = \frac{\pi }{2}\int_0^{\pi /2} {\frac{{\cos x\sin x}}{{{{\cos }^4}x + {{\sin }^4}x}}} $ $dx$

==> $I = \frac{\pi }{4}\int_0^{\pi /2} {\frac{{\tan x\,{{\sec }^2}x}}{{1 + {{\tan }^4}x}}dx} $

Now, Put ${\tan ^2}x = t$, we get

$I = \frac{\pi }{8}\int_0^\infty {\frac{{dt}}{{1 + {t^2}}}= \frac{\pi }{8}[{{\tan }^{ - 1}}t]_0^\infty = \frac{{{\pi ^2}}}{{16}}} $.