To evaluate the integral, apply the trigonometric identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \). The integral transforms to: \[ \int_0^\pi \sin^2(x) \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx. \] This can be separated into: \[ \frac{1}{2} \left[ \int_0^\pi 1 \, dx - \int_0^\pi \cos(2x) \, dx \right]. \] The integral of 1 from 0 to \( \pi \) is: \[ \int_0^\pi 1 \, dx = \pi. \] The integral of \( \cos(2x) \) from 0 to \( \pi \) is: \[ \int_0^\pi \cos(2x) \, dx = 0 \quad \text{(due to the symmetry of \( \cos(2x) \) around \( \pi/2 \))}. \] Therefore, the integral's value is: \[ \frac{1}{2} \times \pi = \frac{\pi}{2}. \] The final result is: \[ \boxed{\frac{\pi}{2}}. \]