Functional equation involving integral

Question

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

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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?

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EDIT

The following solution was given:

Answer 1

As suggested by Jack, the correcte IQ is $$f(x)=x^2+\int_{0}^{\pi/2} \sin(x+t) f(t) dt~~~~(1)$$ $$\implies f(x)=x^2+A \sin x+ B\cos x~~~~(2)$$, such that $$A=\int_{0}^{\pi/2}cos t f(t) dt~~~~(3)$$ $$B=\int_{0}^{\pi/2} \sin t f(t) dt~~~~(4)$$ Inserting (2) in (3) and (4), we get $A,B$ as vreported by the OP.

Answer 2

$$f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$ 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 $$ \left\{ \begin{array}{l} (2 a+\pi (b+4)-8)=4a \\ \pi a+2 b+\pi ^2-8=4b \\ \end{array} \right. $$

$$a= -\frac{\pi ^3-16}{\pi ^2-4},\;b= -\frac{2 \left(3\pi^2-4\pi-8\right)}{\pi ^2-4}$$