The correct answer is $\boxed{\pi}$.
To evaluate the integral, we can use the following trigonometric identity:
$$\sin^2(x) + \cos^2(x) = 1$$
We can then substitute $u = \sin(x)$ to get:
$$\int \sin^2(x) \, dx = \int \frac{1 – \cos^2(x)}{2} \, dx = \frac{x}{2} – \frac{\sin(2x)}{4} + C$$
where $C$ is an arbitrary constant.
We can then substitute back to get:
$$\int \sin^2(x) \, dx = \frac{x}{2} – \frac{\sin(2x)}{4} + C$$
Now, we can evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(x) \, dx$. We get:
$$\begin{align}
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(x) \, dx &= \left[ \frac{x}{2} – \frac{\sin(2x)}{4} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \
&= \frac{\pi}{2} – \frac{\sin(\pi)}{4} – \left( \frac{-\pi}{2} – \frac{\sin(-\pi)}{4} \right) \
&= \frac{\pi}{2} + \frac{\pi}{4} \
&= \pi
\end{align}$$
Therefore, the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( {{\text{x}}\cos {\text{x}}} \right){\text{dx}}$ is $\boxed{\pi}$.
The other options are incorrect because they do not represent the correct value of the integral.