1. The problem is to find the indefinite integral $\int 2 \sin(3 - 7x) \, dx$. 2. We use the formula for integrating sine functions: $\int \sin(ax + b) \, dx = -\frac{1}{a} \cos(ax + b) + C$ where $a$ and $b$ are constants. 3. Here, the integrand is $2 \sin(3 - 7x)$, so we can factor out the constant 2: $$\int 2 \sin(3 - 7x) \, dx = 2 \int \sin(3 - 7x) \, dx$$ 4. Let $u = 3 - 7x$, then $\frac{du}{dx} = -7$ or $dx = -\frac{1}{7} du$. 5. Substitute into the integral: $$2 \int \sin(u) \left(-\frac{1}{7}\right) du = -\frac{2}{7} \int \sin(u) \, du$$ 6. Integrate $\sin(u)$: $$\int \sin(u) \, du = -\cos(u) + C$$ 7. Substitute back: $$-\frac{2}{7} (-\cos(u)) + C = \frac{2}{7} \cos(3 - 7x) + C$$ 8. Therefore, the integral is: $$\int 2 \sin(3 - 7x) \, dx = \frac{2}{7} \cos(3 - 7x) + C$$