Find the Fourier integral of: \[ f(x)=\left\{\begin{array}{lr} \sin (x) & -3 \pi \leq x \leq \pi \\ 0 & x<-3 \pi \text { and } x>\pi \end{array}\right. \]

Step 1 ：Define the function \(f(x) = \sin(x)\) for \(-3\pi \leq x \leq \pi\) and \(f(x) = 0\) for \(x < -3\pi\) and \(x > \pi\).

Step 2 ：Define the Fourier transform integral as \(F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} dx\).

Step 3 ：Substitute the function \(f(x)\) into the Fourier transform integral to get the integrand \(f(x) e^{-ikx} = \sin(x) e^{-ikx}\).

Step 4 ：Calculate the Fourier transform integral from \(-3\pi\) to \(\pi\).

Step 5 ：The Fourier integral of the function \(f(x)\) is given by the expression: \[F(k) = \frac{\sqrt{2}}{2\sqrt{\pi}} \times \left\{\begin{array}{lr} -0.5i\pi e^{i\pi} - 1.5i\pi e^{-3i\pi} & k = -1 \ 1.5i\pi e^{3i\pi} + 0.5i\pi e^{-i\pi} & k = 1 \ -1.0/(k^2 e^{i\pi k} - e^{i\pi k}) + 1.0/(k^2 e^{-3i\pi k} - e^{-3i\pi k}) & \text{otherwise} \end{array}\right.\]

Step 6 ：\(\boxed{F(k) = \frac{\sqrt{2}}{2\sqrt{\pi}} \times \left\{\begin{array}{lr} -0.5i\pi e^{i\pi} - 1.5i\pi e^{-3i\pi} & k = -1 \ 1.5i\pi e^{3i\pi} + 0.5i\pi e^{-i\pi} & k = 1 \ -1.0/(k^2 e^{i\pi k} - e^{i\pi k}) + 1.0/(k^2 e^{-3i\pi k} - e^{-3i\pi k}) & \text{otherwise} \end{array}\right.}\) is the final answer.