The value of `int_(-1//2)^(1//2) |xcos((pix)/(2))|dx` is

Options

A. `(pisqrt(2)+4sqrt(2)-8)/(pi^(2))`
B. `(sqrt(2)+4pisqrt(2)-8)/(pi^(2))`
C. `(pisqrt(2)+4sqrt(2)+8)/(pi^(2))`
D. none of these

Answer

To solve the integral \( I = \int_{-\frac{1}{2}}^{\frac{1}{2}} |x \cos\left(\frac{\pi x}{2}\right)| \, dx \), we will follow these steps:

### Step 1: Analyze the function
The function \( f(x) = |x \cos\left(\frac{\pi x}{2}\right)| \) is defined over the interval \([-1/2, 1/2]\). We need to check if this function is even or odd.

### Step 2: Check if the function is even
We find \( f(-x) \):
\[ f(-x) = |-x \cos\left(\frac{\pi (-x)}{2}\right)| = |x \cos\left(\frac{\pi x}{2}\right)| = f(x) \]
Since \( f(-x) = f(x) \), the function is even.

### Step 3: Use the property of definite integrals
Since \( f(x) \) is even, we can use the property of definite integrals:
\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]
Thus, we have:
\[ I = 2 \int_{0}^{\frac{1}{2}} x \cos\left(\frac{\pi x}{2}\right) \, dx \]

### Step 4: Set up for integration by parts
Let:
- \( u = x \) (then \( du = dx \))
- \( dv = \cos\left(\frac{\pi x}{2}\right) dx \) (then \( v = \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) \))

### Step 5: Apply integration by parts
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int x \cos\left(\frac{\pi x}{2}\right) \, dx = x \cdot \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) - \int \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) \, dx \]

### Step 6: Calculate the integral of \( \sin\left(\frac{\pi x}{2}\right) \)
The integral of \( \sin\left(\frac{\pi x}{2}\right) \) is:
\[ \int \sin\left(\frac{\pi x}{2}\right) \, dx = -\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) \]

### Step 7: Substitute back into the integration by parts formula
Now substituting back:
\[ \int x \cos\left(\frac{\pi x}{2}\right) \, dx = x \cdot \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) + \frac{2}{\pi} \cdot \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) \]
Thus:
\[ \int x \cos\left(\frac{\pi x}{2}\right) \, dx = \frac{2}{\pi} x \sin\left(\frac{\pi x}{2}\right) + \frac{4}{\pi^2} \cos\left(\frac{\pi x}{2}\right) \]

### Step 8: Evaluate from 0 to \( \frac{1}{2} \)
Now we evaluate:
\[ \int_{0}^{\frac{1}{2}} x \cos\left(\frac{\pi x}{2}\right) \, dx = \left[ \frac{2}{\pi} x \sin\left(\frac{\pi x}{2}\right) + \frac{4}{\pi^2} \cos\left(\frac{\pi x}{2}\right) \right]_{0}^{\frac{1}{2}} \]
Calculating at the upper limit \( x = \frac{1}{2} \):
\[ = \frac{2}{\pi} \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4}\right) + \frac{4}{\pi^2} \cos\left(\frac{\pi}{4}\right) \]
\[ = \frac{1}{\pi} \cdot \frac{1}{\sqrt{2}} + \frac{4}{\pi^2} \cdot \frac{1}{\sqrt{2}} = \frac{1 + \frac{4}{\pi}}{\pi \sqrt{2}} \]

At the lower limit \( x = 0 \):
\[ = 0 + \frac{4}{\pi^2} \cdot 1 = \frac{4}{\pi^2} \]

### Step 9: Combine results
Thus:
\[ \int_{0}^{\frac{1}{2}} x \cos\left(\frac{\pi x}{2}\right) \, dx = \frac{1}{\pi \sqrt{2}} + \frac{4}{\pi^2} - 0 \]

### Step 10: Final result
Now, substituting back into the equation for \( I \):
\[ I = 2 \left( \frac{1}{\pi \sqrt{2}} + \frac{4}{\pi^2} \right) = \frac{2}{\pi \sqrt{2}} + \frac{8}{\pi^2} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{2}{\pi \sqrt{2}} + \frac{8}{\pi^2} \]