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.