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Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh 1$. With $k=n\pi$, an antiderivative isThus $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. In the requested factored notation,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.

For period two, the Fourier series convention isSince $f$ is an even function, every $b_n$ is zero. Also $a_0={\int }_{-1}^1\cosh xdx=2\sinh 1$. For $n\ge 1$, an antiderivative useful for the cosine coefficient isEvaluation at $\pm 1$ gives $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. ThusThe periodic extension is continuous at $x=1$, so its Fourier series converges to $\cosh 1$ there. Since $\cos (n\pi )=(-1{)}^n$,Therefore $\boxed{p=2,q=\pi }$, with $q=-\pi$ giving the same expression because only $q^2$ occurs. This is the $\beta =1$ case of Fourier series of periodic hyperbolic sine and cosine.