1. **State the problem:** Evaluate the integral $$\int_{0}^{\frac{\pi}{2}} \frac{\sin^2\left(\frac{\pi}{2}x\right)}{\left(x - \frac{\pi}{2}\right)^2} \, dx.$$ 2. **Analyze the integral:** The integral has a singularity at $x=\frac{\pi}{2}$ because the denominator becomes zero. We need to check if the integral converges and how to handle this singularity. 3. **Rewrite the integral:** Let $a = \frac{\pi}{2}$. Then the integral is $$\int_0^a \frac{\sin^2(ax)}{(x - a)^2} \, dx.$$ 4. **Change variable:** Set $t = a - x$, so when $x=0$, $t=a$, and when $x=a$, $t=0$. The integral becomes $$\int_a^0 \frac{\sin^2(a(a - t))}{(-t)^2} (-dt) = \int_0^a \frac{\sin^2(a^2 - at)}{t^2} \, dt.$$ 5. **Simplify the sine argument:** Since $\sin^2(\theta)$ is periodic with period $\pi$, and $a^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}$, the argument is $a^2 - at = \frac{\pi^2}{4} - \frac{\pi}{2} t$. 6. **Use the identity:** $\sin^2(\alpha) = \frac{1 - \cos(2\alpha)}{2}$. So, $$\sin^2\left(\frac{\pi^2}{4} - \frac{\pi}{2} t\right) = \frac{1 - \cos\left(2 \times \left(\frac{\pi^2}{4} - \frac{\pi}{2} t\right)\right)}{2} = \frac{1 - \cos\left(\frac{\pi^2}{2} - \pi t\right)}{2}.$$ 7. **Rewrite the integral:** $$\int_0^a \frac{\sin^2(a^2 - at)}{t^2} dt = \int_0^a \frac{1 - \cos\left(\frac{\pi^2}{2} - \pi t\right)}{2 t^2} dt.$$ 8. **Use cosine subtraction formula:** $$\cos\left(\frac{\pi^2}{2} - \pi t\right) = \cos\left(\frac{\pi^2}{2}\right) \cos(\pi t) + \sin\left(\frac{\pi^2}{2}\right) \sin(\pi t).$$ 9. **Evaluate constants:** $\cos\left(\frac{\pi^2}{2}\right)$ and $\sin\left(\frac{\pi^2}{2}\right)$ are constants but complicated; however, the integral is improper and difficult to evaluate in closed form. 10. **Conclusion:** The integral is improper due to the singularity at $x=\frac{\pi}{2}$ and involves oscillatory terms. It requires advanced techniques or numerical methods for evaluation. **Final answer:** The integral does not have a simple closed form and is best evaluated numerically or using special functions.