Find and verify solution for $f(x)=a+\int_0^x \sin ({t}) f(x-t) \ dt$ This is the the verification stage of a problem :
$$\text{For $\displaystyle f(x)=a+a\frac{x^2}{2} $, prove that $\displaystyle f(x)=a+\int_0^x \sin ({t})  f(x-t) \ dt$}$$
The problem asked for us to find $f(x)$ to satisfy that condition and once we found it to replace it in the initial condition to prove it's right. And the answer is for certain right, 'cause we solved it in the seminary; but for the life of me, I can't prove it.
I tried integrating by parts until I reached a second degree derivative of $f(x)$ which is $0$, but it doesn't work since I didn't get the initial function.
 A: If you want to verify the identity, we can substitute $f(x)$ and $f(x-t)$ in it getting :
$$\frac{ x^2}{2} = \int_0^x \sin{t}  (1+\frac{(x-t)^2}2)dt=(1+\frac{x^2}{2})\int_0^x\sin t \  dt - x\int_0^x t\sin t \ dt + \frac12\int_0^x t^2 \sin t \ dt $$
You can easily find integrating by parts the last two integrals, so :
$$\int_0^x \sin{t} \ dt   = 1-\cos x$$
$$\int_0^xt \sin t \  dt = \sin x - x \cos x $$
$$\int_0^x t^2 \sin t \ dt  = 2x \sin x -(x^2-2)\cos x -2$$
If you expand the r.h.s. you will obtain the l.h.s.
A: Instead if you want to find the solution I would proceed this way:
substitute $t = x - u$, obtaining
$$\begin{align}f(x)&= a + \int_{0}^{x} \sin(x-u)f(u) \ du \\ & =a+\int_{0}^{x} (\sin(x)\cos(u)-\cos(x)\sin(u))f(u) \ du \\
 &=a+ \sin(x) \int_{0}^{x} \cos(u)f(u) \ du - \cos(x) \int_{0}^{x} \sin(u)f(u) \ du \end{align}$$
So, differentiating the equation you get
$$f'(x) = \cos(x) \int_{0}^{x} cos(u)f(u) \ du +\sin(x) \int_{0}^{x} \sin(u)f(u) \ du $$
and differentiating another time
\begin{align}f''(x)& = f(x) + \cos(x) \int_{0}^{x} \sin(u)f(u) \ du -  \sin(x) \int_{0}^{x} \cos(u)f(u) \ du\\
 &= f(x)+a-f(x) \\& = a \end{align}
Thus, we have $  \ \displaystyle f(x) = a\frac{x^2}{2}+bx+c$, but we need to find the initial conditions: substituting $0$ in the function and its derivative you get  $f(0) = c = a$ and $f'(0) = b = 0$ .
Therefore, we have that  $  \ \displaystyle f(x) = a\frac{x^2}{2}+a \ $ as desidered.
