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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.

Evaluate the integral ∫ sin(6x - 2) dx.

Learn how to evaluate the integral ∫ sin(6x - 2) dx using the substitution method. This solution demonstrates key calculus concepts like integration, trigonometric integrals, and applying the Fundamental Theorem of Calculus. Suitable for high school students in Grades 11-12.

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