integrate((pi-x)*sin(pi*n*x/L)dx) from L/2 to L 

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?

integrate((pi-x)*sin(pi*n*x/L)dx) from L/2 to L

Learn how to solve the definite integral of (π - x) * sin(πnx/L) from L/2 to L. The solution involves distributing terms, applying integration by parts, and simplifying trigonometric integrals. Suitable for advanced high school students or early college level.

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