1. Given

We have

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

and it satisfies

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

We need to find $a+b$.

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2. Expand the kernel

Using

$$\sin(x+y)=\sin x\cos y+\cos x\sin y,$$

we get

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

Let

$$I_1=\int_0^{\pi/2}f(y)\cos y\,dy, \qquad I_2=\int_0^{\pi/2}f(y)\sin y\,dy.$$

Then

$$f(x)=x+I_1\sin x+I_2\cos x.$$

Comparing with the given form

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

we identify

$$I_1=\frac{a}{\pi^2-4}, \qquad I_2=\frac{b}{\pi^2-4}.$$

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3. Use the form of $f(y)$ to compute $I_1,I_2$

Write

$$f(y)=y+I_1\sin y+I_2\cos y.$$

So

$$I_1=\int_0^{\pi/2}f(y)\cos y\,dy =\int_0^{\pi/2}y\cos y\,dy+I_1\int_0^{\pi/2}\sin y\cos y\,dy+I_2\int_0^{\pi/2}\cos^2 y\,dy.$$

Similarly,

$$I_2=\int_0^{\pi/2}f(y)\sin y\,dy =\int_0^{\pi/2}y\sin y\,dy+I_1\int_0^{\pi/2}\sin^2 y\,dy+I_2\int_0^{\pi/2}\sin y\cos y\,dy.$$

Now compute the standard integrals:

$$\int_0^{\pi/2}y\cos y\,dy =\left[y\sin y+\cos y\right]_0^{\pi/2} =\frac{\pi}{2}-1,$$ $$\int_0^{\pi/2}y\sin y\,dy =\left[-y\cos y+\sin y\right]_0^{\pi/2}=1,$$ $$\int_0^{\pi/2}\sin y\cos y\,dy=\frac12,$$ $$\int_0^{\pi/2}\cos^2 y\,dy=\int_0^{\pi/2}\sin^2 y\,dy=\frac{\pi}{4}.$$

Hence,

$$I_1=\left(\frac{\pi}{2}-1\right)+\frac12 I_1+\frac{\pi}{4}I_2$$ $$I_2=1+\frac{\pi}{4}I_1+\frac12 I_2.$$

Rearrange:

$$\frac12 I_1-\frac{\pi}{4}I_2=\frac{\pi}{2}-1$$ $$-\frac{\pi}{4}I_1+\frac12 I_2=1.$$

Multiply by $4$:

$$2I_1-\pi I_2=2\pi-4 \qquad ...(1)$$ $$-\pi I_1+2I_2=4 \qquad ...(2)$$

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4. Solve for $I_1$ and $I_2$

From (1) and (2), in matrix form:

$$\begin{pmatrix} 2 & -\pi\\ -\pi & 2 \end{pmatrix} \begin{pmatrix} I_1\\I_2 \end{pmatrix} = \begin{pmatrix} 2\pi-4\\4 \end{pmatrix}.$$

Since

$$I_1=\frac{a}{\pi^2-4},\qquad I_2=\frac{b}{\pi^2-4},$$

substitute into (1), (2):

$$2\frac{a}{\pi^2-4}-\pi\frac{b}{\pi^2-4}=2\pi-4$$ $$-\pi\frac{a}{\pi^2-4}+2\frac{b}{\pi^2-4}=4.$$

Multiply both equations by $(\pi^2-4)$:

$$2a-\pi b=(2\pi-4)(\pi^2-4) \qquad ...(3)$$ $$-\pi a+2b=4(\pi^2-4). \qquad ...(4)$$

But we only need $a+b$. So solve directly for $I_1+I_2$ first.

Add (1) and (2):

$$(2-\pi)(I_1+I_2)=2\pi.$$

Thus

$$I_1+I_2=\frac{2\pi}{2-\pi}=-\frac{2\pi}{\pi-2}.$$

Now

$$a+b=(\pi^2-4)(I_1+I_2).$$

So

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

Using

$$\pi^2-4=(\pi-2)(\pi+2),$$

we get

$$a+b=-(\pi-2)(\pi+2)\cdot \frac{2\pi}{\pi-2} =-2\pi(\pi+2).$$

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5. Match with options

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

which is Option A.

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6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.