1. Problem: Evaluate the indefinite integral $$\int \sin(2x)\,dx$$. 2. Use the substitution rule: let $$u = 2x$$, then $$\frac{du}{dx} = 2$$ or $$dx = \frac{du}{2}$$. 3. Rewrite the integral in terms of $$u$$: $$\int \sin(2x)\, dx = \int \sin(u) \cdot \frac{du}{2} = \frac{1}{2} \int \sin(u)\, du$$. 4. Integrate $$\sin(u)$$ with respect to $$u$$: $$\int \sin(u)\, du = -\cos(u) + C$$. 5. Substitute back $$u = 2x$$: $$\frac{1}{2} \int \sin(u)\, du = \frac{1}{2}(-\cos(u)) + C = -\frac{1}{2} \cos(2x) + C$$. 6. Final answer: $$\int \sin(2x)\, dx = -\frac{1}{2} \cos(2x) + C$$.