To solve the given integral \( \pi^2 \int_0^2 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx \), where \([x]\) denotes the greatest integer function (GIF), we can break the integral into two parts based on the intervals defined by the GIF.
### Step-by-Step Solution:
1. **Split the Integral**:
Since the greatest integer function \([x]\) changes at integer values, we split the integral from \(0\) to \(2\) into two parts:
\[
\int_0^2 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx = \int_0^1 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx + \int_1^2 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx
\]
2. **Evaluate the First Integral**:
For \(x\) in the interval \([0, 1)\), \([x] = 0\). Thus, \((x - [x])^{[x]} = (x - 0)^0 = 1\).
\[
\int_0^1 \sin\left(\frac{\pi}{2} x\right) \cdot 1 \, dx = \int_0^1 \sin\left(\frac{\pi}{2} x\right) \, dx
\]
The integral of \(\sin(kx)\) is \(-\frac{1}{k} \cos(kx)\). Therefore:
\[
\int_0^1 \sin\left(\frac{\pi}{2} x\right) \, dx = -\frac{2}{\pi} \left[\cos\left(\frac{\pi}{2} x\right)\right]_0^1 = -\frac{2}{\pi} \left(\cos\left(\frac{\pi}{2}\right) - \cos(0)\right) = -\frac{2}{\pi} (0 - 1) = \frac{2}{\pi}
\]
3. **Evaluate the Second Integral**:
For \(x\) in the interval \([1, 2)\), \([x] = 1\). Thus, \((x - [x])^{[x]} = (x - 1)^1 = x - 1\).
\[
\int_1^2 \sin\left(\frac{\pi}{2} x\right) (x - 1) \, dx
\]
We can use integration by parts here. Let \(u = x - 1\) and \(dv = \sin\left(\frac{\pi}{2} x\right) dx\). Then \(du = dx\) and \(v = -\frac{2}{\pi} \cos\left(\frac{\pi}{2} x\right)\).
\[
\int (x - 1) \sin\left(\frac{\pi}{2} x\right) \, dx = \left(-\frac{2}{\pi} (x - 1) \cos\left(\frac{\pi}{2} x\right)\right) \bigg|_1^2 + \frac{2}{\pi} \int \cos\left(\frac{\pi}{2} x\right) \, dx
\]
Evaluating the boundary terms:
\[
-\frac{2}{\pi} \left[(2 - 1) \cos(\pi) - (1 - 1) \cos\left(\frac{\pi}{2}\right)\right] = -\frac{2}{\pi} \left[1 \cdot (-1) - 0\right] = \frac{2}{\pi}
\]
Now, integrating \(\cos\left(\frac{\pi}{2} x\right)\):
\[
\int \cos\left(\frac{\pi}{2} x\right) \, dx = \frac{2}{\pi} \sin\left(\frac{\pi}{2} x\right) + C
\]
Thus:
\[
\frac{2}{\pi} \left[\frac{2}{\pi} \sin\left(\frac{\pi}{2} x\right)\right] \bigg|_1^2 = \frac{4}{\pi^2} \left[\sin(\pi) - \sin\left(\frac{\pi}{2}\right)\right] = \frac{4}{\pi^2} \left[0 - 1\right] = -\frac{4}{\pi^2}
\]
4. **Combine the Results**:
Now, we can combine the results of both integrals:
\[
\int_0^2 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx = \frac{2}{\pi} + \left(\frac{2}{\pi} - \frac{4}{\pi^2}\right) = \frac{4}{\pi} - \frac{4}{\pi^2}
\]
5. **Final Calculation**:
Finally, multiplying by \(\pi^2\):
\[
\pi^2 \left(\frac{4}{\pi} - \frac{4}{\pi^2}\right) = 4\pi - 4 = 4(\pi - 1)
\]
### Final Answer:
\[
\text{The value of } \pi^2 \int_0^2 \sin\left(\frac{\pi}{2} x\right) (x - [x])^{[x]} \, dx = 4(\pi - 1)
\]
