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The equation provided is:

\[
a_n = \frac{2}{1} \int_{-1}^{1} f(t) \cos(nt) \, dt = \int_{-1}^{0} (t+1)\cos(nt) \, dt + \int_{0}^{1} (-t+1)\cos(nt) \, dt
\]

Let's solve each integral step by step.

### Step 1: Evaluate the first integral
The first integral is:

\[
\int_{-1}^{0} (t+1)\cos(nt) \, dt
\]

We will use integration by parts to solve this integral. Let:

- \( u = t+1 \), hence \( du = dt \)
- \( dv = \cos(nt) \, dt \), hence \( v = \frac{\sin(nt)}{n} \)

Using integration by parts, \( \int u \, dv = uv - \int v \, du \), we get:

\[
\int (t+1)\cos(nt) \, dt = (t+1)\frac{\sin(nt)}{n} - \int \frac{\sin(nt)}{n} \, dt
\]

The integral of \( \sin(nt) \) is:

\[
\int \sin(nt) \, dt = -\frac{\cos(nt)}{n}
\]

Thus:

\[
\int_{-1}^{0} (t+1)\cos(nt) \, dt = \left[\frac{(t+1)\sin(nt)}{n} + \frac{\cos(nt)}{n^2}\right]_{-1}^{0}
\]

Now, evaluate from \( -1 \) to \( 0 \):

At \( t = 0 \):

\[
\frac{(0+1)\sin(0)}{n} + \frac{\cos(0)}{n^2} = 0 + \frac{1}{n^2} = \frac{1}{n^2}
\]

At \( t = -1 \):

\[
\frac{(-1+1)\sin(-n)}{n} + \frac{\cos(-n)}{n^2} = 0 + \frac{\cos(n)}{n^2} = \frac{\cos(n)}{n^2}
\]

So, the first integral evaluates to:

\[
\frac{1}{n^2} - \frac{\cos(n)}{n^2} = \frac{1 - \cos(n)}{n^2}
\]

### Step 2: Evaluate the second integral
The second integral is:

\[
\int_{0}^{1} (-t+1)\cos(nt) \, dt
\]

Similarly, using integration by parts:

- \( u = -t+1 \), \( du = -dt \)
- \( dv = \cos(nt) \, dt \), \( v = \frac{\sin(nt)}{n} \)

Thus:

\[
\int (-t+1)\cos(nt) \, dt = (-t+1)\frac{\sin(nt)}{n} - \int \frac{\sin(nt)}{n} \, (-dt)
\]

This simplifies to:

\[
\int_{0}^{1} (-t+1)\cos(nt) \, dt = \left[\frac{(-t+1)\sin(nt)}{n} - \frac{\cos(nt)}{n^2}\right]_{0}^{1}
\]

Now, evaluate from \( 0 \) to \( 1 \):

At \( t = 1 \):

\[
\frac{(-1+1)\sin(n)}{n} - \frac{\cos(n)}{n^2} = 0 - \frac{\cos(n)}{n^2} = -\frac{\cos(n)}{n^2}
\]

At \( t = 0 \):

\[
\frac{(0-1)\sin(0)}{n} - \frac{\cos(0)}{n^2} = 0 - \frac{1}{n^2} = -\frac{1}{n^2}
\]

So, the second integral evaluates to:

\[
-\frac{\cos(n)}{n^2} + \frac{1}{n^2} = \frac{1 - \cos(n)}{n^2}
\]

### Step 3: Combine the results
Now, summing the results of the two integrals:

\[
a_n = \frac{1 - \cos(n)}{n^2} + \frac{1 - \cos(n)}{n^2} = \frac{2(1 - \cos(n))}{n^2}
\]

So the final answer is:

\[
a_n = \frac{2(1 - \cos(n))}{n^2}
\]

Would you like further clarification or details on any of these steps?

Here are some related questions to further explore:

1. How does the result change if \( n \) is a specific integer value, like \( n = 1 \) or \( n = 2 \)?
2. What happens if the cosine function is replaced with a sine function in the integrals?
3. How can we apply this method to different piecewise functions?
4. What is the physical interpretation of \( a_n \) in the context of Fourier series?
5. How does the Fourier series change if the function \( f(t) \) has different boundary conditions?

**Tip:** When dealing with trigonometric integrals, integration by parts is a powerful tool, especially when the product of a polynomial and a trigonometric function is involved.

Explore how to calculate Fourier series coefficients using trigonometric integrals step-by-step. This problem involves advanced concepts in Fourier series and integration techniques. Suitable for advanced undergraduate students and researchers.

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