an = 2/$$ \pi $$ $$ {\displaylines\int_0^{\pi}\!x.\cos x.\cos\left(\text{nx}\right)\,\mathrm{d}x\\ \placeholder{}\\ } $$

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?

---

**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.

an = 2/$$ \pi $$ $$ {\displaylines\int_0^{\pi}\!x.\cos x.\cos\left(\text{nx}\right)\,\mathrm{d}x\ }$$

Explore how to solve a trigonometric integral involving cos(x) and cos(nx) using the product-to-sum identity and integration by parts. This step-by-step solution is ideal for advanced high school students and undergraduates, covering key calculus and trigonometry concepts.

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