1. Simplify the integrand

We need to evaluate

$$I=\int_{-\pi}^{\pi} \left|\pi-|x|\right|\,dx.$$

For $x\in[-\pi,\pi]$, we have

$$|x|\le \pi \implies \pi-|x|\ge 0.$$

So the outer modulus is unnecessary:

$$\left|\pi-|x|\right|=\pi-|x|.$$

Hence,

$$I=\int_{-\pi}^{\pi}(\pi-|x|)\,dx.$$

2. Use symmetry

The function $\pi-|x|$ is even, so

$$I=2\int_0^{\pi}(\pi-x)\,dx.$$

3. Evaluate the integral

$$\int_0^{\pi}(\pi-x)\,dx n=\left[\pi x-\frac{x^2}{2}\right]_0^{\pi} =\pi^2-\frac{\pi^2}{2} =\frac{\pi^2}{2}.$$

Therefore,

$$I=2\cdot \frac{\pi^2}{2}=\pi^2.$$

4. Match with the options

$$\int_{-\pi}^{\pi} \left|\pi-|x|\right|dx=\pi^2.$$

So the correct option is:

• A: $\pi^2$