Area of ascending regions on implicit plot - $\cos(y^y) = \sin(x)$ Take the equation
$$ \cos \left( y^y \right) = \sin(x) $$
The plot forms a kind of skew checkerboard, where each "tile" shrinks as $y$ increases. I was attempting to find the sum of the areas (the blue regions in the image below), but cannot find an expression for each area. I would like to find an expression where $a_n$ is the $n$-th region up and show whether $\sum a_n$ diverges or converges and if so the value of the sum.

I'm not entirely sure how to find the corners of the region algebraically, but the first one has approximate corners: left- $\left(-\frac{\pi}{2},2.47 \right)$, top- $ \left(\frac{\pi}{2},2.62 \right)$, right- $ \left(\frac{3\pi}{2},2.47 \right)$, bottom- $ \left(\frac{\pi}{2},2.26 \right)$. If we approximate the shape of the first region as a rhombus, we get $$A \approx \frac{(3\pi/2 + \pi/2) (2.62 - 2.26)}{2} = 1.130973 \text{ or } a_0 \approx 1.130973.$$
 A: Disclaimer: This is only a partial answer.
The area of the region you are referring to is
$$
A = \sum_{n=1}^\infty\int_{y_n}^{y_{n+1}}\left[\left(\frac{3\pi}2-\arcsin\cos(y^y)\right)-\left(\frac{\pi}2+\arcsin\cos(y^y)\right)\right]\text{d} y
$$
where $y_n$ is the sequence of successive points where the function $\arcsin\cos(y^y)$ is not differentiable (we will determine these later). The way we have obtained this integral is by noting that the eqaution $\cos(y^y) = \sin(x)$ is a periodic continuation of the equation $x=\arcsin(\cos(y^y))$ (and its reflection across the line $x=\pi/2$), with period $\pi$.
Note: You can see this by going ahead and graphing the above function of $y$ next to your graph of the implicit equation above.
EDIT: I can't believe I missed it but I just realized that the sum simplifies to $\int_{y_1}^\infty$ of the function. You can skip some of the sections below about $y_n$ if you prefer.
Let $f(y) = \arcsin(cos(y^y))$. The reason we're integrating over $[3\pi/2-f(y)] - [\pi/2+f(y)]$ as opposed to simply $[\pi-f(y)]-f(y)$ is because $f(y)$ is negative, with minimum value $\pi/2$, so we shift everything up to make the area positive.
Let $I_n$ denote the integral in the summand above. Then
$$
I_n = \int_{y_n}^{y_{n+1}}(\pi - 2f(y))\text{d} y = \pi(y_{n+1}-y_n)-2\int_{y_n}^{y_{n+1}}f(y)\text{d} y
$$
Wolfram Alpha gives no closed form for the indefinite integral of $f(y)$ in terms of elementary functions (or transcendental/special ones for that matter), so it may be of use to us to know what exactly the $y_n$ are.
What exactly is happening at the points where $f(y)$ is not differentiable? These are the points where $\cos(y^y)$ reaches a critical point, and then goes in the opposite direction. Another way of seeing this is noting that $\arcsin\cos x$ is a transformation of the triangle wave, which has inflection points at the critical points of $\cos x$.
Hence, are points $y_n$ are the critical points of $\cos(y^y)$. The derivative of $\cos(y^y)$ is
$$
\frac{\text{d}}{\text{d}y}\cos(y^y) = -y^y(\ln(y)+1)\sin(y^y)
$$
Setting this equal to zero gives $y_1 = \frac1e$ (which gives $\ln(y)=-1$), and $y_n = \exp(\text{W}(\ln(\pi n)))$ for $n>1$ (where $\text{W}(x)$ is the Lambert W function).
Consider $n>1$. Then
$$
I_n = \int_{e^{\text{W}(\ln(\pi n))}}^{e^{\text{W}(\ln(\pi(n+1)))}}\arcsin(\cos(x^x))\text{d}x
$$
We can make the substitution $u=x^x$ by noting that $x = \text{W}(\ln u)$, and so $\text{d}u = x^x(\ln x + 1)\text{d} x = u(\text{W}(\ln u) + 1)\text{d}x$, which gives
$$
I_n = \int_{\pi n}^{\pi (n+1)}\frac{\arcsin(\cos u)}{\text{W}(\ln u) + 1}\frac{\text{d}u}{u}
$$
This is where I've run out of time, and partially out of ideas. I don't see any simple route to take to evaluate this integral, and I haven't taken a good look at its asymptotics to determine if the sum will even converge. I hope this has at least shed some light on an approach to the problem.
EDIT: I realized that I might've been overcomplicating things by introducing $y_n$, and not noting that since the integrals in the summand have consecutive start and end points, we can write the area as
$$
\begin{aligned}
A & = \int_{y_1}^\infty(\pi-2f(y))\text{d}y \\
& = \lim_{N\to\infty}\left(\pi\left(N-\frac1e\right)-2\int_{e^{-1/e}}^{N^N}\frac{\arcsin(\cos(u))}{\text{W}(\ln u)+1}\frac{\text{d}u}{u}\right)
\end{aligned}
$$
I haven't looked much into this limit yet.
