$\int_{0}^{\frac{\pi}{2}} \sin^{2}{x} dx = ?$

To solve the integral $\int_{0}^{\frac{\pi}{2}} \sin^{2}{x} \, dx$, we can use the trigonometric identity for $\sin^2{x}$:

$\sin^2{x} = \frac{1 - \cos{2x}}{2}$

Substitute this identity into the integral:

$\int_{0}^{\frac{\pi}{2}} \sin^{2}{x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos{2x}}{2} \, dx$

This separates into two integrals:

$= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos{2x} \, dx$

Calculate each integral:

$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{1}{2} \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \times \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$

$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos{2x} \, dx = \frac{1}{2} \left[ \frac{\sin{2x}}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{4} \left( \sin{(\pi)} - \sin{0} \right) = \frac{1}{4} \times (0 - 0) = 0$

Thus, the integral evaluates to:

$\frac{\pi}{4} - 0 = \frac{\pi}{4}$

Therefore, $\int_{0}^{\frac{\pi}{2}} \sin^{2}{x} \, dx = \frac{\pi}{4} \text{ (Ans.)}$