For a positive integer n, let `I _(n) int _(-pi)^(pi) ((pi)/(2) -|x|) cos nx dx`
Find the value of `[I _(1) + I _(3) +I_(4)]` where [.] denotes greatest integer function.

Answer

To solve the given problem, we need to evaluate the integral

\[ I_n = \int_{-\pi}^{\pi} \left( \frac{\pi}{2} - |x| \right) \cos(nx) \, dx \]

for \( n = 1, 3, 4 \) and then find the value of \( \lfloor I_1 + I_3 + I_4 \rfloor \).

### Step 1: Determine the nature of the function \( f(x) = \frac{\pi}{2} - |x| \cos(nx) \)

Since \( f(-x) = f(x) \), the function is even. Therefore, we can simplify the integral:

\[ I_n = 2 \int_0^{\pi} \left( \frac{\pi}{2} - x \right) \cos(nx) \, dx \]

### Step 2: Split the integral

Now we can split the integral into two parts:

\[ I_n = 2 \left( \int_0^{\pi} \frac{\pi}{2} \cos(nx) \, dx - \int_0^{\pi} x \cos(nx) \, dx \right) \]

### Step 3: Evaluate the first integral

The first integral is:

\[ \int_0^{\pi} \frac{\pi}{2} \cos(nx) \, dx = \frac{\pi}{2} \left[ \frac{\sin(nx)}{n} \right]_0^{\pi} = \frac{\pi}{2} \cdot \frac{\sin(n\pi)}{n} = 0 \]

since \( \sin(n\pi) = 0 \).

### Step 4: Evaluate the second integral using integration by parts

For the second integral, we use integration by parts:

Let \( u = x \) and \( dv = \cos(nx) \, dx \).

Then, \( du = dx \) and \( v = \frac{\sin(nx)}{n} \).

Using integration by parts:

\[ \int x \cos(nx) \, dx = x \cdot \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} \, dx \]

Calculating the second integral:

\[ \int \sin(nx) \, dx = -\frac{\cos(nx)}{n} \]

Thus,

\[ \int x \cos(nx) \, dx = x \cdot \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \]

Evaluating from \( 0 \) to \( \pi \):

\[ \left[ x \cdot \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \right]_0^{\pi} = \left( \pi \cdot \frac{\sin(n\pi)}{n} + \frac{\cos(n\pi)}{n^2} \right) - \left( 0 + \frac{1}{n^2} \right) \]

This simplifies to:

\[ 0 + \frac{\cos(n\pi)}{n^2} - \frac{1}{n^2} = \frac{\cos(n\pi) - 1}{n^2} \]

### Step 5: Combine results for \( I_n \)

Now substituting back into our expression for \( I_n \):

\[ I_n = 2 \left( 0 - \left( \frac{\cos(n\pi) - 1}{n^2} \right) \right) = \frac{2(1 - \cos(n\pi))}{n^2} \]

### Step 6: Calculate \( I_1, I_3, I_4 \)

1. For \( n = 1 \):
\[ I_1 = \frac{2(1 - \cos(\pi))}{1^2} = \frac{2(1 - (-1))}{1} = \frac{2 \cdot 2}{1} = 4 \]

2. For \( n = 3 \):
\[ I_3 = \frac{2(1 - \cos(3\pi))}{3^2} = \frac{2(1 - (-1))}{9} = \frac{2 \cdot 2}{9} = \frac{4}{9} \]

3. For \( n = 4 \):
\[ I_4 = \frac{2(1 - \cos(4\pi))}{4^2} = \frac{2(1 - 1)}{16} = 0 \]

### Step 7: Calculate \( I_1 + I_3 + I_4 \)

Now we can find:

\[ I_1 + I_3 + I_4 = 4 + \frac{4}{9} + 0 = 4 + \frac{4}{9} = \frac{36}{9} + \frac{4}{9} = \frac{40}{9} \]

### Step 8: Find the greatest integer function

Finally, we need to find:

\[ \lfloor I_1 + I_3 + I_4 \rfloor = \lfloor \frac{40}{9} \rfloor = \lfloor 4.4444 \rfloor = 4 \]

Thus, the final answer is:

\[ \boxed{4} \]