1. We are asked to integrate the function $$\int \cos(6x+4)\, dx.$$\n\n2. Recall that the integral of $$\cos(u)$$ with respect to $$u$$ is $$\sin(u) + C$$, where $$C$$ is the constant of integration.\n\n3. Since the argument inside the cosine is $$6x+4$$, use substitution: let $$u = 6x + 4$$, then $$\frac{du}{dx} = 6$$ or $$dx = \frac{du}{6}$$.\n\n4. Substitute these into the integral:\n$$\int \cos(6x+4)\, dx = \int \cos(u) \cdot \frac{du}{6} = \frac{1}{6} \int \cos(u)\, du.$$\n\n5. Integrate with respect to $$u$$:\n$$\frac{1}{6} \int \cos(u)\, du = \frac{1}{6} \sin(u) + C.$$\n\n6. Substitute back for $$u$$ to express the answer in terms of $$x$$:\n$$\frac{1}{6} \sin(6x+4) + C.$$\n\nTherefore, the integral is $$\boxed{\frac{\sin(6x+4)}{6} + C}$$.\n\nAmong the choices given, option d matches this result.