To solve this integral:

$\int \sin(6x - 2) \, dx$

we can use substitution. Let's go through the steps:

1. Substitute: Let $u = 6x - 2$. Then, $\frac{du}{dx} = 6$ or $dx = \frac{du}{6}$.

2. Rewrite the integral: Substitute $u$ and $dx$ into the integral: $\int \sin(6x - 2) \, dx = \int \sin(u) \cdot \frac{du}{6} = \frac{1}{6} \int \sin(u) \, du.$

3. Integrate: The integral of $\sin(u)$ is $-\cos(u)$: $\frac{1}{6} \int \sin(u) \, du = -\frac{1}{6} \cos(u) + C.$

4. Substitute back $u = 6x - 2$: $-\frac{1}{6} \cos(6x - 2) + C.$

So the answer is:

$\boxed{-\frac{1}{6} \cos(6x - 2) + C}$

This matches option C in your answer choices.

Would you like more details or have any other questions?

Here are some related questions to consider:

1. How does substitution work in integrals, and why is it useful?
2. Why is the derivative of $\cos(u)$ equal to $-\sin(u)$?
3. What are other methods to evaluate trigonometric integrals?
4. How does integration differ from differentiation for trigonometric functions?
5. What are common mistakes to avoid in trigonometric integrals?

Tip: Always check if a simple substitution can simplify an integral before proceeding with more complex techniques.