$\int\limits_0^{\frac{\pi}{4}} \frac{\sin 2 x}{\cos ^4 x+\sin ^4 x} d x=$

$\frac{\pi}{4}$

The correct answer is Option (2) → $\frac{\pi}{4}$

$\int\limits_0^{\frac{\pi}{4}} \frac{\sin 2 x}{\cos ^4 x+\sin ^4 x} d x=\int\frac{2\sin x\cos x}{\cos ^4 x+\sin ^4 x}dx$

Use substitution $t = \tan x$. 

$\rightarrow$ Integral from 0 to 1 of $2t / (1 + t^4)$

Substitute $u = t^2$.

$\rightarrow$ Integral of $1 / (1 + u^2)$ from 0 to 1

$\rightarrow \tan^{-1}(1) - \tan^{-1}(0) = \pi/4$