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 $
- DNone of these
$ = 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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