1. **Stating the problem:** Calculate the integral $$\int \sin x \, dx$$ using integration by parts. 2. **Recall the formula for integration by parts:** $$\int u \, dv = uv - \int v \, du$$ where you choose parts of the integrand as $u$ and $dv$. 3. **Choose parts:** Let $u = 1$ (a constant) and $dv = \sin x \, dx$. Then, $$du = 0 \, dx$$ and $$v = -\cos x$$ 4. **Apply the formula:** $$\int \sin x \, dx = uv - \int v \, du = 1 \cdot (-\cos x) - \int (-\cos x) \cdot 0 \, dx = -\cos x - 0 = -\cos x + C$$ 5. **Final answer:** $$\int \sin x \, dx = -\cos x + C$$ This is a basic integral and can also be found directly from standard integral tables.