To simplify the integral, we apply the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \): \[ \int_0^\pi \sin^2(x) \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx. \] The integral is then separated: \[ \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^\pi 1 \, dx - \frac{1}{2} \int_0^\pi \cos(2x) \, dx. \] The first part evaluates to: \[ \int_0^\pi 1 \, dx = \pi. \] The second part evaluates to: \[ \int_0^\pi \cos(2x) \, dx = 0 \quad \text{(due to the symmetry of \( \cos(2x) \) around \( \pi/2 \))}. \] Substituting these values, the integral simplifies to: \[ \frac{1}{2} \times \pi = \frac{\pi}{2}. \] The final result is: \[ \boxed{\frac{\pi}{2}}. \]