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 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 \( [I_1 + I_3 + I_4] \), where \( [.] \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Understanding the Integral**:
The integral \( I_n \) is defined as:
\[ I_n = \int_{-\pi}^{\pi} \left( \frac{\pi}{2} - |x| \right) \cos(nx) \, dx \]

2. **Recognizing the Even Function**:
Since \( |x| \) is an even function and \( \cos(nx) \) is also even, the product \( \left( \frac{\pi}{2} - |x| \right) \cos(nx) \) is even. Therefore, we can simplify the integral:
\[ I_n = 2 \int_{0}^{\pi} \left( \frac{\pi}{2} - x \right) \cos(nx) \, dx \]

3. **Expanding the Integral**:
We can split the integral:
\[ I_n = 2 \left( \int_{0}^{\pi} \frac{\pi}{2} \cos(nx) \, dx - \int_{0}^{\pi} x \cos(nx) \, dx \right) \]

4. **Calculating the First Integral**:
The first integral can be calculated as follows:
\[ \int_{0}^{\pi} \frac{\pi}{2} \cos(nx) \, dx = \frac{\pi}{2} \left[ \frac{\sin(nx)}{n} \right]_{0}^{\pi} = \frac{\pi}{2} \cdot \frac{\sin(n\pi)}{n} = 0 \]
(since \( \sin(n\pi) = 0 \) for any integer \( n \)).

5. **Calculating 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 \( v = \frac{\sin(nx)}{n} \).
Thus,
\[ \int x \cos(nx) \, dx = x \cdot \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} \, dx \]
Evaluating the second integral:
\[ \int \sin(nx) \, dx = -\frac{\cos(nx)}{n} \]
Therefore,
\[ \int x \cos(nx) \, dx = x \cdot \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \]
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:
\[ 0 + \frac{\cos(n\pi)}{n^2} - \frac{1}{n^2} = \frac{\cos(n\pi) - 1}{n^2} \]

6. **Putting It All Together**:
Now substituting back into \( I_n \):
\[ I_n = 2 \left( 0 - \frac{\cos(n\pi) - 1}{n^2} \right) = \frac{2(1 - \cos(n\pi))}{n^2} \]

7. **Calculating \( I_1, I_3, I_4 \)**:
- For \( n = 1 \):
\[ I_1 = \frac{2(1 - \cos(\pi))}{1^2} = \frac{2(1 - (-1))}{1} = 4 \]
- For \( n = 3 \):
\[ I_3 = \frac{2(1 - \cos(3\pi))}{3^2} = \frac{2(1 - (-1))}{9} = \frac{4}{9} \]
- For \( n = 4 \):
\[ I_4 = \frac{2(1 - \cos(4\pi))}{4^2} = \frac{2(1 - 1)}{16} = 0 \]

8. **Summing Up**:
Now we find \( I_1 + I_3 + I_4 \):
\[ I_1 + I_3 + I_4 = 4 + \frac{4}{9} + 0 = 4 + \frac{4}{9} = \frac{36}{9} + \frac{4}{9} = \frac{40}{9} \]

9. **Finding the Greatest Integer**:
Finally, we take the greatest integer:
\[ \left[ I_1 + I_3 + I_4 \right] = \left[ \frac{40}{9} \right] = 4 \]

### Final Answer:
\[ \boxed{4} \]