1. Stated problem: Calculate the integral $$\int (1 - x) \sin x \, dx$$. 2. Formula and rules: Use integration by parts, where $$\int u \, dv = uv - \int v \, du$$. 3. Choose $$u = 1 - x$$ and $$dv = \sin x \, dx$$. 4. Compute derivatives and integrals: $$du = -dx$$ $$v = -\cos x$$ 5. Apply integration by parts: $$\int (1 - x) \sin x \, dx = (1 - x)(-\cos x) - \int -\cos x (-dx)$$ 6. Simplify the integral: $$= -(1 - x) \cos x - \int \cos x \, dx$$ 7. Integrate $$\cos x$$: $$\int \cos x \, dx = \sin x$$ 8. Final expression: $$-(1 - x) \cos x - \sin x + C$$ 9. Simplify: $$-\cos x + x \cos x - \sin x + C$$ Answer: $$\int (1 - x) \sin x \, dx = x \cos x - \cos x - \sin x + C$$.