.Let `f(x)=x+(a)/(x^(2)-4)sin x+(b)/(pi^(2)-4)cos x,x in R` be a function which satisfies `f(x)=x+int_(0)^( pi/2)sin(x+y)f(y)dy`.Then `(a+b)` is equal to

Options

A. `-pi(pi+2)`
B. `-pi(pi-2)`
C. `-2 pi(pi+2)`
D. `-2 pi(pi-2)`

Answer

To solve the problem, we start with the given function:

\[ f(x) = x + \frac{a}{x^2 - 4} \sin x + \frac{b}{\pi^2 - 4} \cos x \]

and the equation:

\[ f(x) = x + \int_0^{\frac{\pi}{2}} \sin(x+y) f(y) \, dy \]

### Step 1: Expand the integral

Using the identity for sine, we can expand \(\sin(x+y)\):

\[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]

Thus, we can rewrite the integral:

\[ \int_0^{\frac{\pi}{2}} \sin(x+y) f(y) \, dy = \int_0^{\frac{\pi}{2}} (\sin x \cos y + \cos x \sin y) f(y) \, dy \]

### Step 2: Substitute \(f(y)\)

Substituting \(f(y)\) into the integral gives:

\[ \int_0^{\frac{\pi}{2}} (\sin x \cos y + \cos x \sin y) \left( y + \frac{a}{y^2 - 4} \sin y + \frac{b}{\pi^2 - 4} \cos y \right) dy \]

### Step 3: Split the integral

This can be split into three separate integrals:

1. \(\int_0^{\frac{\pi}{2}} y \sin x \cos y \, dy\)
2. \(\int_0^{\frac{\pi}{2}} y \cos x \sin y \, dy\)
3. The terms involving \(\sin y\) and \(\cos y\).

### Step 4: Evaluate the integrals

1. The first integral can be evaluated as:

\[ \int_0^{\frac{\pi}{2}} y \sin x \cos y \, dy = \sin x \int_0^{\frac{\pi}{2}} y \cos y \, dy \]

Using integration by parts, we find:

\[ \int_0^{\frac{\pi}{2}} y \cos y \, dy = \left[ y \sin y \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \sin y \, dy = 0 - [ -\cos y ]_0^{\frac{\pi}{2}} = 1 \]

Thus, the first integral becomes \(\sin x\).

2. The second integral:

\[ \int_0^{\frac{\pi}{2}} y \cos x \sin y \, dy = \cos x \int_0^{\frac{\pi}{2}} y \sin y \, dy \]

Using integration by parts again:

\[ \int_0^{\frac{\pi}{2}} y \sin y \, dy = \left[ -y \cos y \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos y \, dy = 0 + 1 = 1 \]

Thus, this integral becomes \(\cos x\).

### Step 5: Combine results

Combining these results, we have:

\[ f(x) = x + \sin x + \cos x + \text{(terms involving a and b)} \]

### Step 6: Set coefficients equal

From the original function \(f(x)\), we can equate coefficients for \(\sin x\) and \(\cos x\):

- Coefficient of \(\sin x\): \(\frac{a}{x^2 - 4} = 1\)
- Coefficient of \(\cos x\): \(\frac{b}{\pi^2 - 4} = 1\)

### Step 7: Solve for \(a\) and \(b\)

From the equations:

1. \(a = x^2 - 4\)
2. \(b = \pi^2 - 4\)

### Step 8: Calculate \(a + b\)

Now we find \(a + b\):

\[ a + b = (x^2 - 4) + (\pi^2 - 4) = x^2 + \pi^2 - 8 \]

Since \(x\) is arbitrary and does not affect the constants \(a\) and \(b\), we can conclude that:

\[ a + b = \pi^2 - 8 \]

### Final Result

Thus, the value of \(a + b\) is:

\[ \boxed{0} \]