$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^{-4} x \mathrm{~d} x=$

The integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{-4} x dx$ is improper due to discontinuity at $x = 0$ where $\sin x = 0$. Evaluation shows divergence as $\lim_{x \to 0^{+}} -\left(\cot x + \frac{\cot^{3} x}{3}\right) = -\infty$ and similarly from the left.

Assuming the intended integral was $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{4} x dx$ due to common transcription errors and finite options given:

The even function property yields $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{4} x dx = 2\int_{0}^{\frac{\pi}{4}} \sec^{4} x dx$.
Using $\sec^{2} x = 1 + \tan^{2} x$ and substitution $u = \tan x$, $du = \sec^{2} x dx$:

$2\int_{0}^{1} (1 + u^{2}) du = 2\left[u + \frac{u^{3}}{3}\right]_{0}^{1} = 2\left(1 + \frac{1}{3}\right) = \frac{8}{3}$

This matches Option A.