We are asked to evaluate the definite integral $\int_0^\pi \sin^2 x \, dx$.

1. Problem Description

We are asked to evaluate the definite integral $\int_0^\pi \sin^2 x \, dx$.

2. Solution Steps

We will use the following trigonometric identity to simplify the integrand:

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

So, the integral becomes:

$\int_0^\pi \sin^2 x \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx$

$= \frac{1}{2} \int_0^\pi (1 - \cos(2x)) \, dx$

$= \frac{1}{2} \left[ \int_0^\pi 1 \, dx - \int_0^\pi \cos(2x) \, dx \right]$

$= \frac{1}{2} \left[ x \Big|_0^\pi - \frac{1}{2}\sin(2x) \Big|_0^\pi \right]$

$= \frac{1}{2} \left[ (\pi - 0) - \frac{1}{2}(\sin(2\pi) - \sin(0)) \right]$

$= \frac{1}{2} \left[ \pi - \frac{1}{2}(0 - 0) \right]$

$= \frac{1}{2} \pi = \frac{\pi}{2}$.

3. Final Answer

The final answer is $\frac{\pi}{2}$.