\(\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}(\sin x)^{-4} \mathrm{~d} x\) has the value

Let
\(\begin{aligned}
& \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\sin x)^{-4} \mathrm{~d} x \\
&=2 \int_0^{\frac{\pi}{4}}(\sin x)^{-4} \mathrm{~d} x \\
& \quad \ldots\left[\because(\sin x)^{-4} \text { is an even function }\right] \\
&=2 \int_0^{\frac{\pi}{4}} \frac{1}{\sin ^4 x} \mathrm{~d} x \\
&=2 \int_0^{\frac{\pi}{4}} \frac{\sec ^4 x}{\tan ^4 x} \mathrm{~d} x \\
&=2 \int_0^{\frac{\pi}{4}} \frac{\left(1+\tan ^2 x\right) \sec ^2 x}{\tan ^4 x} \mathrm{~d} x
\end{aligned}\)
Let \(\tan x=\mathrm{t}\) when \(x=0\), we get \(\mathrm{t}=0\)
when \(x=\frac{\pi}{4}\), we get \(\mathrm{t}=1\)
Also, \(\sec ^2 x \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad \mathrm{I} & =2 \int_0^1 \frac{1+\mathrm{t}^2}{\mathrm{t}^4} \mathrm{dt} \\
& =2\left[\int_0^1 \frac{1}{\mathrm{t}^4} \mathrm{dt}+\int_0^1 \frac{1}{\mathrm{t}^2} \mathrm{dt}\right] \\
& =2\left[\frac{\left[\mathrm{t}^{-3}\right]_0^1}{-3}+\frac{\left[\mathrm{t}^{-1}\right]_0^1}{-1}\right] \\
& =\frac{-8}{3}
\end{aligned}\)