The integral provided is as follows:

$a_n = \frac{2}{\pi} \int_0^{\pi} x \cos(x) \cos(nx) \, \mathrm{d}x$

Let's solve this step by step.

Step 1: Use product-to-sum identities for the cosine terms

We can use the product-to-sum identity for cosine functions to simplify the product of $\cos(x)$ and $\cos(nx)$. The identity is:

$\cos(A)\cos(B) = \frac{1}{2} \left( \cos(A-B) + \cos(A+B) \right)$

Substituting $A = x$ and $B = nx$, we get:

$\cos(x)\cos(nx) = \frac{1}{2} \left( \cos((n-1)x) + \cos((n+1)x) \right)$

Thus, the integral becomes:

$a_n = \frac{2}{\pi} \int_0^{\pi} x \cdot \frac{1}{2} \left( \cos((n-1)x) + \cos((n+1)x) \right) \, \mathrm{d}x$

Simplifying:

$a_n = \frac{1}{\pi} \int_0^{\pi} x \left( \cos((n-1)x) + \cos((n+1)x) \right) \, \mathrm{d}x$

Step 2: Split the integral

Now, we can split the integral into two parts:

$a_n = \frac{1}{\pi} \left( \int_0^{\pi} x \cos((n-1)x) \, \mathrm{d}x + \int_0^{\pi} x \cos((n+1)x) \, \mathrm{d}x \right)$

Step 3: Solve each integral

The integral we need to solve is of the form:

$I_m = \int_0^{\pi} x \cos(mx) \, \mathrm{d}x$

This integral can be solved using integration by parts. Let’s compute this integral next:

1. Choose $u = x$ and $dv = \cos(mx) \, dx$.
2. Then, $du = dx$ and $v = \frac{\sin(mx)}{m}$.

Now, applying integration by parts:

$I_m = \left[ \frac{x \sin(mx)}{m} \right]_0^{\pi} - \int_0^{\pi} \frac{\sin(mx)}{m} \, dx$

The boundary terms become:

$\left[ \frac{x \sin(mx)}{m} \right]_0^{\pi} = \frac{\pi \sin(m\pi)}{m} - \frac{0 \sin(0)}{m} = 0$

Since $\sin(m\pi) = 0$, the first part vanishes.

Now, for the remaining integral:

$I_m = -\frac{1}{m} \int_0^{\pi} \sin(mx) \, dx$

The integral of $\sin(mx)$ is:

$\int \sin(mx) \, dx = -\frac{\cos(mx)}{m}$

Thus,

$I_m = -\frac{1}{m} \left[ -\frac{\cos(mx)}{m} \right]_0^{\pi} = \frac{1}{m^2} \left( \cos(m\pi) - \cos(0) \right)$

We know that $\cos(m\pi) = (-1)^m$ and $\cos(0) = 1$, so:

$I_m = \frac{1}{m^2} \left( (-1)^m - 1 \right)$

Step 4: Apply this result to our integrals

For the first integral, $m = n-1$:

$I_{n-1} = \frac{1}{(n-1)^2} \left( (-1)^{n-1} - 1 \right)$

For the second integral, $m = n+1$:

$I_{n+1} = \frac{1}{(n+1)^2} \left( (-1)^{n+1} - 1 \right)$

Step 5: Combine the results

Substituting the results of the integrals back into the expression for $a_n$, we get:

$a_n = \frac{1}{\pi} \left( \frac{1}{(n-1)^2} \left( (-1)^{n-1} - 1 \right) + \frac{1}{(n+1)^2} \left( (-1)^{n+1} - 1 \right) \right)$

This is the simplified expression for $a_n$.

Would you like further clarification or details on any part of this process?

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5 related questions:

1. How does the product-to-sum identity work for trigonometric functions?
2. Can you explain the steps of integration by parts in more detail?
3. How do boundary conditions affect definite integrals?
4. Why does $\sin(m\pi)$ equal zero in certain integrals?
5. How would the solution change if a different trigonometric identity was used?

Tip: When using integration by parts, it's important to carefully choose $u$ and $dv$ to simplify the integral efficiently.