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

Question

Aicharya · Accepted Answer

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

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

Solution