<s>[INST] Are there any elementary functions $\beta(x)$ that follows this integral $\int_{y-1}^{y} \beta(x) dx =\cos(y)$ Are there any simple functions $\beta(x)$ that follows this integral $$\int_{y-1}^{y} \beta(x) dx =\cos(y)$$
I think there is an infinite amount of solutions that are continuous everywhere but how can I find one that only uses elementary functions?
 [/INST] Let $f(y) :=\beta(y) - (A\sin y + B \cos y) = \beta(y-1) - (A\sin (y-1) + B\cos (y-1))$
then $(A+1)\sin y + B\cos y =A\sin(y-1) +B \cos(y-1) $.
$=(A\cos 1 + B \sin 1)\sin y + (B \cos 1 - A\sin 1)\cos y$
$\therefore A+1 = (A\cos 1 + B \sin 1) \; , \; B = (B \cos 1 - A\sin 1) $
$A+1 = \frac{1}{A}((A^2+B^2)\cos 1 - B^2) $  ( $\because B(\cos 1 - 1) = A \sin 1 $)
$\Rightarrow (A^2 + B^2)(\cos 1 -1) = A , (A^2+B^2)\sin 1 = B$
Let $A^2 + B^2 = R$. then $R = R^2 ( 2 - 2\cos 1) $ ,  $R= \frac{1}{2-2\cos 1}$
so $ A = -1/2$ , $ B = \frac{\sin 1}{2(1-\cos 1)}$.
Now, $f(y)$ is periodic. ( period of $f(y)$ is $1$)
and
$ \int_{y-1}^{y} f(x)dx = \cos y - \int_{y-1}^{y} (A\sin x +B \cos x)dx \\ =\cos y + A(\cos y - \cos (y-1)) - B(\sin y - \sin (y-1))$.
because period of $f(y)$ is $1$, $\int_{y-1}^{y} f(x) dx = \int_{0}^{1} f(x) dx$
$= \cos1 + (A\cos 1 - A)-B(\sin 1) =\frac{1}{2} + \frac{1}{2} \cos 1 - \frac{\sin^2 1}{2(1-\cos 1)}$.
Finally, we get $ \beta(y) = -\frac{1}{2}\sin y + \frac{\sin 1}{2(1 - \cos 1)} \cos y + (\frac{1}{2} + \frac{1}{2} \cos 1 - \frac{\sin^2 1}{2(1-\cos 1)} ) $
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