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

The integral to evaluate is $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|) dx$

Since $|x|$ changes at $x = 0$, the integral splits into two parts:

For $-\frac{\pi}{2} \leq x < 0$: $|x| = -x$

For $0 \leq x \leq \frac{\pi}{2}$: $|x| = x$

The integral becomes:

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

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Using the properties $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$:

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

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For the first integral:

$\int_{-\frac{\pi}{2}}^{0} (-\sin x + \cos x) dx = [\cos x + \sin x]_{-\frac{\pi}{2}}^{0}$

At $x = 0$: $\cos(0) + \sin(0) = 1 + 0 = 1$

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

Result: $1 - (-1) = 2$

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For the second integral:

$\int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) dx = [-\cos x + \sin x]_{0}^{\frac{\pi}{2}}$

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

At $x = 0$: $-\cos(0) + \sin(0) = -1 + 0 = -1$

Result: $1 - (-1) = 2$

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Total value:

$2 + 2 = 4$

Therefore, the value of the integral is $4$.