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The expression you have uploaded involves two integrals and appears to be a part of solving a specific mathematical or physics-related problem, likely related to Fourier series or boundary value problems.

The expression can be broken down as follows:

\[
\frac{2}{l} \left[ \int_{0}^{l/2} \frac{2k}{l} x \cos\left( \frac{n \pi}{l} x \right) dx + \int_{l/2}^{l} \frac{2k}{l} (l - x) \cos\left( \frac{n \pi}{l} x \right) dx \right]
\]

Let's solve each integral step by step.

### First Integral:
\[
I_1 = \int_{0}^{l/2} \frac{2k}{l} x \cos\left( \frac{n \pi}{l} x \right) dx
\]

We will need to use integration by parts for this integral.

### Second Integral:
\[
I_2 = \int_{l/2}^{l} \frac{2k}{l} (l - x) \cos\left( \frac{n \pi}{l} x \right) dx
\]

Again, integration by parts will be used for solving this.

I'll go ahead and start solving these integrals step-by-step for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, if you'd like, I can guide you through solving this step-by-step manually. Would you like me to proceed with that, or would you like to try again later when the tool is available?

Here are some related questions you might consider while we wait:

1. Do you have any boundary conditions associated with this integral problem?
2. How would you approach the integration by parts for each of these integrals?
3. Are you familiar with trigonometric integral techniques in such problems?
4. What physical or mathematical context is this problem from (e.g., wave equations or Fourier analysis)?
5. Have you considered any symmetry in the problem that might simplify the integration?

**Tip**: When solving integrals involving trigonometric functions, often the use of integration by parts and identities can simplify the process significantly.

Step-by-step solution to the integral involving trigonometric cosine function and linear components. This problem explores advanced integration techniques such as integration by parts and Fourier series. Suitable for undergraduate students.

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