1. The problem is to find the indefinite integral $$\int \left(2\sin(x+2) + 3\sin(x+3)\right) dx.$$\n\n2. We use the linearity of the integral: $$\int (a f(x) + b g(x)) dx = a \int f(x) dx + b \int g(x) dx,$$ where $a$ and $b$ are constants.\n\n3. Applying this, we get: $$2 \int \sin(x+2) dx + 3 \int \sin(x+3) dx.$$\n\n4. Recall the integral formula: $$\int \sin(u) du = -\cos(u) + C.$$\n\n5. For each integral, let $u = x + c$ where $c$ is a constant. Then $du = dx$, so: $$\int \sin(x+c) dx = -\cos(x+c) + C.$$\n\n6. Substitute back: $$2(-\cos(x+2)) + 3(-\cos(x+3)) + C = -2\cos(x+2) - 3\cos(x+3) + C.$$\n\n7. Therefore, the final answer is $$\boxed{-2\cos(x+2) - 3\cos(x+3) + C}.$$