MCQ
$\int_{\; - \pi }^\pi {\frac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\;dx} = $
  • A
    $\pi /4$
  • B
    $\pi /2$
  • C
    $3\pi /2$
  • $\pi $

Answer

Correct option: D.
$\pi $
d
(d) $I = \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)$

Adding $(i)$ and $(ii)$ we get,

$2I = 4\int_0^{\pi /2} {dx = 4 \times \frac{\pi }{2} = 2\pi } $

==> $I = \pi $.

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