Let $$f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$$, $$x \in \mathbb{R}$$ be a function which satisfies $$f(x) = x + \int_0^{\pi/2} \sin(x+y)f(y)dy$$. Then $$(a+b)$$ is equal to

Let the given function be:

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

We are given that $$f(x)$$ satisfies the integral equation:

$$f(x) = x + \int_0^{\pi/2} \sin(x+y)f(y)dy$$

Step 1: Expand the trigonometric identity inside the integral

Using $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$, we can rewrite the integral equation as:

$$f(x) = x + \sin x \int_0^{\pi/2} \cos y f(y)dy + \cos x \int_0^{\pi/2} \sin y f(y)dy$$

Comparing this with the given definition of $$f(x)$$, we get:

$$\frac{a}{\pi^2-4} = \int_0^{\pi/2} \cos y f(y)dy$$

$$\frac{b}{\pi^2-4} = \int_0^{\pi/2} \sin y f(y)dy$$

Step 2: Substitute $$f(y)$$ into the two integral equations

Let $$k = \pi^2 - 4$$. Then $$f(y) = y + \frac{a}{k}\sin y + \frac{b}{k}\cos y$$.

For the first equation:

$$\frac{a}{k} = \int_0^{\pi/2} \cos y \left(y + \frac{a}{k}\sin y + \frac{b}{k}\cos y\right) dy$$

$$\frac{a}{k} = \int_0^{\pi/2} y\cos y \, dy + \frac{a}{k}\int_0^{\pi/2} \sin y\cos y \, dy + \frac{b}{k}\int_0^{\pi/2} \cos^2 y \, dy$$

Evaluating each component:

* $$\int_0^{\pi/2} y\cos y \, dy = [y\sin y]_0^{\pi/2} - \int_0^{\pi/2} \sin y \, dy = \frac{\pi}{2} - 1$$

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

* $$\int_0^{\pi/2} \cos^2 y \, dy = \frac{\pi}{4}$$

Substituting these back into the equation for $$a$$:

$$\frac{a}{k} = \left(\frac{\pi}{2} - 1\right) + \frac{a}{2k} + \frac{b\pi}{4k}$$

$$\frac{a}{2k} - \frac{b\pi}{4k} = \frac{\pi - 2}{2}$$

$$2a - b\pi = 2k(\pi - 2)$$

Since $$k = \pi^2 - 4 = (\pi-2)(\pi+2)$$, we have:

$$2a - b\pi = 2(\pi-2)(\pi+2)(\pi-2) = 2(\pi-2)^2(\pi+2)$$

Step 3: Solve for the second equation

$$\frac{b}{k} = \int_0^{\pi/2} \sin y \left(y + \frac{a}{k}\sin y + \frac{b}{k}\cos y\right) dy$$

$$\frac{b}{k} = \int_0^{\pi/2} y\sin y \, dy + \frac{a}{k}\int_0^{\pi/2} \sin^2 y \, dy + \frac{b}{k}\int_0^{\pi/2} \sin y\cos y \, dy$$

Evaluating each component:

* $$\int_0^{\pi/2} y\sin y \, dy = [-y\cos y]_0^{\pi/2} + \int_0^{\pi/2} \cos y \, dy = 1$$

* $$\int_0^{\pi/2} \sin^2 y \, dy = \frac{\pi}{4}$$

* $$\int_0^{\pi/2} \sin y\cos y \, dy = \frac{1}{2}$$

Substituting these back into the equation for $$b$$:

$$\frac{b}{k} = 1 + \frac{a\pi}{4k} + \frac{b}{2k}$$

$$\frac{b}{2k} - \frac{a\pi}{4k} = 1$$

$$2b - a\pi = 4k = 4(\pi^2 - 4)$$

Step 4: Solve the system of linear equations for $$a$$ and $$b$$

Solving the linear equations simultaneously yields:

$$a = 8 - 8\pi$$

$$b = -2\pi^2 + 4\pi - 8$$

Step 5: Calculate $$(a+b)$$

$$(a+b) = (8 - 8\pi) + (-2\pi^2 + 4\pi - 8)$$

$$(a+b) = -2\pi^2 - 4\pi$$

Final Answer:

$$(a+b) = -2\pi^2 - 4\pi$$