= Solution
Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh1$. With $k=n\pi$, an <antiderivative> is
$$
\int\cosh x\cos(kx)dx=\frac{\sinh x\cos(kx)+k\cosh x\sin(kx)}{1+k^2}.
$$
Thus $a_n=2(-1)^n\sinh1/(1+n^2\pi^2)$. In the requested factored notation,
$$
\boxed{A_0=1,\qquad A_n(x)=\frac{2(-1)^n\cos(n\pi x)}{1+n^2\pi^2}.}
$$
This is the period-two specialization of the <Fourier series of periodic hyperbolic sine and cosine>. The function values match at the endpoints, although its derivative has a jump in the periodic extension.
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