1. The problem is to evaluate the integral $$\int \sin x \, dx$$. 2. The formula for the integral of sine is $$\int \sin x \, dx = -\cos x + C$$, where $C$ is the constant of integration. 3. Applying the formula directly, we get: $$\int \sin x \, dx = -\cos x + C$$ 4. This means the antiderivative of $\sin x$ is $-\cos x$ plus a constant. 5. Therefore, the final answer is: $$-\cos x + C$$