MCQ
$\int_0^{\pi /2} {\frac{{x\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}} \,dx = $
- A$0$
- B$\frac{\pi }{8}$
- C$\frac{{{\pi ^2}}}{8}$
- ✓$\frac{{{\pi ^2}}}{{16}}$
$ = \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}}} $.
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