We are tasked with evaluating the definite integral:

$I = \int_{L/2}^{L} \left( \pi - x \right) \sin \left( \frac{\pi n x}{L} \right) \, dx$

Step 1: Distribute the terms

First, we distribute the $(\pi - x)$ term over the $\sin\left( \frac{\pi n x}{L} \right)$ function:

$I = \int_{L/2}^{L} \pi \sin \left( \frac{\pi n x}{L} \right) \, dx - \int_{L/2}^{L} x \sin \left( \frac{\pi n x}{L} \right) \, dx$

We will handle each of these integrals separately.

Step 2: Solve the first integral

Let's first solve the integral:

$I_1 = \int_{L/2}^{L} \pi \sin \left( \frac{\pi n x}{L} \right) \, dx$

The integral of $\sin \left( \frac{\pi n x}{L} \right)$ with respect to $x$ is:

$\int \sin \left( \frac{\pi n x}{L} \right) \, dx = -\frac{L}{\pi n} \cos \left( \frac{\pi n x}{L} \right)$

Thus,

$I_1 = \pi \left[ -\frac{L}{\pi n} \cos \left( \frac{\pi n x}{L} \right) \right]_{L/2}^{L}$

Evaluating this from $x = L/2$ to $x = L$:

$I_1 = -\frac{L}{n} \left( \cos \left( \pi n \right) - \cos \left( \frac{\pi n}{2} \right) \right)$

Since $\cos(\pi n) = (-1)^n$, we get:

$I_1 = -\frac{L}{n} \left( (-1)^n - \cos \left( \frac{\pi n}{2} \right) \right)$

Step 3: Solve the second integral

Now, let's solve the second integral:

$I_2 = \int_{L/2}^{L} x \sin \left( \frac{\pi n x}{L} \right) \, dx$

We can solve this using integration by parts. Let:

• $u = x$ (so $du = dx$)
• $dv = \sin \left( \frac{\pi n x}{L} \right) \, dx$ (so $v = -\frac{L}{\pi n} \cos \left( \frac{\pi n x}{L} \right)$)

Then, applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$I_2 = -\frac{L}{\pi n} \left[ x \cos \left( \frac{\pi n x}{L} \right) \right]_{L/2}^{L} + \frac{L}{\pi n} \int_{L/2}^{L} \cos \left( \frac{\pi n x}{L} \right) \, dx$

First, evaluate the boundary terms:

$\left[ x \cos \left( \frac{\pi n x}{L} \right) \right]_{L/2}^{L} = L \cos (\pi n) - \frac{L}{2} \cos \left( \frac{\pi n}{2} \right)$

Thus,

$I_2 = -\frac{L}{\pi n} \left( L (-1)^n - \frac{L}{2} \cos \left( \frac{\pi n}{2} \right) \right)$

Next, the remaining integral:

$\int_{L/2}^{L} \cos \left( \frac{\pi n x}{L} \right) \, dx = \frac{L}{\pi n} \left[ \sin \left( \frac{\pi n x}{L} \right) \right]_{L/2}^{L}$

Evaluating:

$\sin \left( \frac{\pi n x}{L} \right) \Big|_{L/2}^{L} = \sin(\pi n) - \sin \left( \frac{\pi n}{2} \right)$

Since $\sin (\pi n) = 0$, this term becomes:

$\int_{L/2}^{L} \cos \left( \frac{\pi n x}{L} \right) \, dx = - \frac{L}{\pi n} \sin \left( \frac{\pi n}{2} \right)$

So, the final expression for $I_2$ becomes:

$I_2 = -\frac{L}{\pi n} \left( L (-1)^n - \frac{L}{2} \cos \left( \frac{\pi n}{2} \right) \right) + \frac{L^2}{\pi^2 n^2} \sin \left( \frac{\pi n}{2} \right)$

Step 4: Combine the results

Now, combine $I_1$ and $I_2$:

$I = I_1 - I_2$

After substituting the expressions for $I_1$ and $I_2$, we get the final result. You can simplify this expression based on the specific value of $n$ to get a numerical result.

Would you like to see the final result for a specific $n$ or any further simplification?