Let $I=\int\limits_{-\pi / 4}^{\pi / 4} \sin ^{-4} x d x$
$=\int\limits_{-\pi / 4}^{\pi / 4} \frac{1}{\sin ^{4} x} d x$
$=\int\limits_{-\pi / 4}^{\pi / 4} \text{cosec}^{4} x d x=\int_{-\pi / 4}^{\pi / 4}\left(\text{cosec}^{2} x\right)^{2} d x$
$=\int\limits_{-\pi / 4}^{\pi / 4}\left(1+\cot ^{2} x\right) \text{cosec}^{2} x d x$
Put $\cot x=t$
$\Rightarrow -\text{cosec}^{2} x d x=d t$
$\therefore I=-\int\limits_{-1}^{1}\left(1+t^{2}\right) d t=-2 \int\limits_{0}^{1}\left(1+t^{2}\right) d t$
$=-2\left[t+\frac{t^{3}}{3}\right]_{0}^{1}$
$=-2\left[1+\frac{1}{3}\right]$
$=-\frac{8}{3}$