MCQ
$\int_0^{\pi /4} {\frac{{4\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} = $
  • A
    $\pi /4$
  • B
    $\pi /2$
  • $\pi $
  • D
    None of these

Answer

Correct option: C.
$\pi $
c
(c) $4\int_0^{\pi /4} {\frac{{\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }} = 4\int_0^{\pi /4} {\frac{{2\sin \theta \cos \theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} } $

$ = 4\int_0^{\pi /4} {\frac{{2\tan \theta \,{{\sec }^2}\theta \,d\theta }}{{{{\tan }^4}\theta + 1}}} $

{Dividing numerator and denominator by ${\cos ^4}\theta $ }

Now put ${\tan ^2}\theta = t $

$\Rightarrow 2\tan \theta \,{\sec ^2}\theta \,d\theta = dt$,  then the reduced form is 

$4\int_0^1 {\frac{{dt}}{{{t^2} + 1}}} = 4[{\tan ^{ - 1}}t]_0^1 = 4\left[ {\frac{1}{4}\pi - 0} \right] = \pi $.

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