Area bounded by $\cos x+\cos y=1$

Write the implicit equation into parametric form: $$\begin{cases} x(t)=t\\ y(t)=\pm\arccos(1-\cos t) \end{cases}\quad t\in [-\dfrac{\pi}2,\dfrac{\pi}2]\tag{1}$$

The curve looks like:

Use Green's theorem,

$${\rm Area}=4\int_0^{\tfrac{\pi}2}x(t){\rm d}y(t)=4\int_0^{\tfrac{\pi}2}\dfrac{-t \;\sin t\; {\rm d}t}{\sqrt{(2-\cos t) \cos t}}\tag{2}$$

Substitute $u=\cos t$, $t=\arccos u$:

$${\rm Area}=4\int_0^1 \dfrac{\arccos u}{\sqrt{2 u-u^2}}{\rm d}u \tag{3}$$

Then calculate (3) in Mathematica you can obtain the desired output:

(1/(9 Sqrt[[Pi]]))8 (9 Gamma[3/4]^2 HypergeometricPFQ[{1/4, 1/4, 3/4, 3/4}, {1/2, 5/4, 5/4}, 1/4] + Gamma[5/4]^2 HypergeometricPFQ[{3/4, 3/4, 5/4, 5/4}, {3/2, 7/4, 7/4}, 1/ 4])