Question
$\int_{\; - \pi }^\pi {\frac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\;dx} = $
$\therefore $$I = 2 \times 2\int_0^{\pi /2} {\frac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\;dx} $.....$(i)$
$I = 4\int_0^{\pi /2} {\frac{{{{\sin }^4}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^4}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^4}\left( {\frac{\pi }{2} - x} \right)}}\;dx} $
$I = 4\int_0^{\pi /2} {\frac{{{{\cos }^4}x}}{{{{\cos }^4}x + {{\sin }^4}x}}\;dx} $.....$(ii)$
$(i)$ व $(ii)$ को जोडने पर,
$2I = 4\int_0^{\pi /2} {dx = 4 \times \frac{\pi }{2} = 2\pi } $
==> $I = \pi $.
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