Take $I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$

Let $2 x=t$

$2 d x=d t$

$d x=\frac{d t}{2}$

$\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+\cos ^4 t} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right)}{\sin ^4 t+\cos ^4 t}\end{aligned}$

$\begin{aligned} & 2 I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\frac{\pi}{2}}{\sin ^4 t+\cos ^4 t} d t \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{d t}{\sin ^4 t+\cos ^4 t} \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{\sec ^4 t}{1+\tan ^4 t} d t\end{aligned}$

$\begin{aligned} & \text { Put tant }=y \\\\ & \sec ^2 t d t=d y \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+y^2\right) d y}{1+y^4} \\\\ & =\frac{\pi}{8} \int\limits_0^{\infty} \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2} d y \\\\ & = \frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+\frac{1}{y^2}\right) d y}{2+\left(y-\frac{1}{y}\right)^2}\end{aligned}$

$\begin{aligned} & \text { Put, } y-\frac{1}{y}=4 \\\\ & 2 I=\frac{\pi}{8} \int\limits_{-\infty}^{\infty} \frac{d u}{2+u^2} \\\\ & =\frac{\pi}{8 \sqrt{2}}\left[\tan ^{-1} \frac{y}{\sqrt{2}}\right]_{-\infty}^{\infty} \\\\ & =\frac{\sqrt{2} \pi^2}{32}\end{aligned}$