Functional equation involving integral 
Find $f(x)$ given that it satisfies the equation $f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+1) dt$


My attempt:
Let $\int_0^{\pi/2}f(t)dt=c$
So
$$f(x)=x^2+c\sin(x+1)$$
Integrate w.r.t. $x$ from $0$ to $\pi/2$
$$\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}x^2dx+c\int_0^{\pi/2}\sin(x+1)dx$$
$$c=\frac{x^3}{3}\bigg|_0^{\frac{\pi}{2}}-c(\cos(x+1))\bigg|_0^{\frac{\pi}{2}}$$
$$c=\frac{\pi^3}{24}+c(\sin(1)+\cos(1))$$
$$\implies c=\frac{\pi^3}{24(1-\sin1-\cos1)}$$
Plugging it back we obtain:
$$f(x)=x^2+\frac{\pi^3\sin(x+1)}{24(1-\sin1-\cos1)}$$
This however doesn't seem to match with the given answer:
$$f(x) = x^2-\frac {2(3\pi^2-4\pi-8)}{\pi^2-4}\sin x-\frac {\pi^3-16}{\pi^2-4} \cos x$$
I have tried expanding $\sin(x+1)$ in my expression but it didn't lead to anything.
Are these expressions equivalent? If yes, how do I inter-convert them?

EDIT
The following solution was given:


$$f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
differentiate twice wrt $x$
$$f''(x) = 2-\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
but
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt=f(x)-x^2$$
thus we get
$$f''(x)=2-f(x)+x^2$$
$$f(x) = x^2+a \cos (x)+b \sin (x)$$
Compute now
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt=\frac{1}{4} \left(\left(\pi  a+2 b+\pi ^2-8\right) \sin (x)+(2 a+\pi  (b+4)-8) \cos (x)\right)$$
we know that $f(x)-x^2=a \cos (x)+b \sin (x)$ so we have
$$a \cos (x)+b \sin (x)=\frac{1}{4} \left(\left(\pi  a+2 b+\pi ^2-8\right) \sin (x)+(2 a+\pi  (b+4)-8) \cos (x)\right)$$
which leads to the system of two linear equations
$$
\left\{
\begin{array}{l}
 (2 a+\pi  (b+4)-8)=4a \\
 \pi  a+2 b+\pi ^2-8=4b \\
\end{array}
\right.
$$
$$a= -\frac{\pi ^3-16}{\pi ^2-4},\;b= -\frac{2 \left(3\pi^2-4\pi-8\right)}{\pi ^2-4}$$
