The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)dx$, is equal to:

Since $\sin|{-x}| = \sin|x|$ and $\cos|{-x}| = \cos|x|$, the integrand $(\sin|x| + \cos|x|)$ is an even function.

For any even function integrated over a symmetric interval $[-a, a]$:

$\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$

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$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx = 2\int_{0}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx$

For $x \in [0,\, \pi/2]$, $x$ is non-negative, so $|x| = x$:

$= 2\int_{0}^{\frac{\pi}{2}} (\sin x + \cos x)\,dx$

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$= 2\Big[-\cos x + \sin x\Big]_{0}^{\frac{\pi}{2}}$

At $x = \dfrac{\pi}{2}$:

$-\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = -(0) + (1) = 1$

At $x = 0$:

$-\cos 0 + \sin 0 = -(1) + (0) = -1$

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$= 2\Big[(1) - (-1)\Big]$

$= 2 \times 2$

$= \boxed{4}$