1. The problem is to evaluate the definite integral $$\int_0^\pi \sin x \, dx$$. 2. The formula for the integral of sine is $$\int \sin x \, dx = -\cos x + C$$, where $C$ is the constant of integration. 3. For definite integrals, we use the Fundamental Theorem of Calculus: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $F$ is an antiderivative of $f$. 4. Applying this to our problem, we find an antiderivative of $\sin x$ which is $-\cos x$. 5. Evaluate at the bounds: $$-\cos \pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$$. 6. Therefore, the value of the integral is $$2$$.