To integrate the function \( y = 2\cos(6x) \), we follow the standard integration rules for trigonometric functions.

### Solution:
The integral we need to compute is:

\[
\int 2 \cos(6x) \, dx
\]

We can factor out the constant \( 2 \) and focus on integrating \( \cos(6x) \):

\[
2 \int \cos(6x) \, dx
\]

Now, using the fact that:

\[
\int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C
\]

where \( k = 6 \), we have:

\[
\int \cos(6x) \, dx = \frac{1}{6} \sin(6x) + C
\]

Thus, the integral becomes:

\[
2 \times \left( \frac{1}{6} \sin(6x) \right) + C = \frac{1}{3} \sin(6x) + C
\]

### Final Answer:
\[
\int 2 \cos(6x) \, dx = \frac{1}{3} \sin(6x) + C
\]

Would you like further details or explanations?

### Related Questions:
1. How would you integrate \( y = 3\sin(4x) \)?
2. What is the integral of \( \cos(kx) \) in general?
3. Can you differentiate the result to verify the integration?
4. How would the integration change if it was \( y = 2 \cos(6x) + 5x \)?
5. How does the presence of a constant multiplier inside the cosine function affect the integration?

### Tip:
Always check your integration by differentiating the result—this helps ensure you haven't made any mistakes.

This guide explains how to integrate the function y = 2cos(6x) using basic integration techniques. It covers key trigonometric concepts and integration rules, providing a clear step-by-step solution for students in Grades 10-12.

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