Finding the area surrounded by the part of the implicit equation $\sin (y^x) = \cos (x^y)$ such that $y\le 2n-x$ where $n$ is the solution to $n^n=\frac{\pi}{4}$ where $n<0.5$ bounded by the $x$ and $y$ axes.
This problem has no context because I am trying to solve it just for fun.
First, I knew what part of the implicit equation I wanted to find the area of (the part that looks similar to a lemniscate) but I did not know how to properly define the integral because I realized I could not find the bounds of this part of the equation easily.
I then manipulated the equation $\sin (y^x) = \cos (x^y)$ into $y^x = \sin^{-1}(\cos (x^y))$ which I then realized I could simplify it to $y^{x}+x^{y}=\frac{\pi}{2}$.
I am currently struggling to find the bounds to set up the integral which I will then hopefully be able use to find the area. I have tried using the implicit derivative but to no avail.
My only success so far was to find the point where the slope is $-1$ which is $(n,n)$ where $n$ is the solution to $n^n=\frac{\pi}{4}$ where $n<0.5$ although I am not sure if this information is actually helpful.
Here is my graph on Desmos if anyone wants to see it: https://www.desmos.com/calculator/x3lfyatuxj
Thanks for the help in advance
Edit: While randomly trying things in the calculator, I found that the equation $y=ex$ goes through the local maximum.
I can plug this in to the original equation to get $x^{ex}+e^{x}x^{x}=\frac{\pi}{2}$.
I thought I would be able to easily find x but I was unable to and WolframAlpha just gave the approximation of $x\approx0.0499491320167935$. It also gives a solution of $x\approx0.508081340341750$ but it is outside of the original bounds $y\le 2n-x$.
$\def\E{\operatorname E}$ Partial answer
Applying Lagrange reversion, a Stirling S2 $s_n^{(m)}$ differentiation formula, and factorial power $n^{(m)}$:
$$x^y+y^x=\frac\pi2\implies y=\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}}{dy^{n-1}}\left(\frac\pi2-x^y\right)^\frac nx\right|_0=\sum_{n=1}^\infty\sum_{k=0}^n\left(\frac nx\right)^{(k)}\frac{s_{n-1}^{(k)}(-1)^k}{n!}\left(\frac\pi2-1\right)^{\frac nx-k} \ln^{n-1}(x)$$
but this is hard to integrate, so we expand. Each sum can go to $\infty$ and, numerically, can be switched. Next, convert to the Pochhammer symbol $(n)_m$, $a^{(k)}=(a-k+1)_k$, and use its Stirling S1 $S_n^{(m)}$ expansion. Also, we switch sums to apply $\frac{d^nx^m}{dm^n}=x^m\ln^n(x)$:
$$x^y+y^x=\frac\pi2\implies y=\sum_{m=0}^\infty\sum_{n=1}^\infty\sum_{k=0}^n(-1)^{k+n+1}n^m\frac{S_k^{(m)}s_{n-1}^{(k)}}{n!}\left(\frac\pi2-1\right)^{\frac nx-k}\frac{d^{n-1}}{dm^{n-1}}x^{-m}$$
shown here. $y(x)$ is the branch bounding region $2$.
The actual maximum of $x^y+y^x=\frac\pi2$ is where $\frac d{dx}\left(x^y+y^x-\frac\pi2\right)=0\iff yx^y+xy^x\ln(y)=0$ intersects it at $(a,b)=(0.0500360\ldots,0.1357758\ldots)$. The Lagrange reversion series requires increasingly more terms as $x\to a$ for accuracy, but does not diverge, so it is plausible to assume the series works on the entire branch. Therefore, region $3$’s area uses the En function:
$$\begin{aligned}\int_0^a y(x)dx=\int_0^b \sum_{m=0}^\infty\sum_{n=1}^\infty\sum_{k=0}^n(-1)^{k+n+1}n^m\frac{S_k^{(m)}s_{n-1}^{(k)}}{n!}\left(\frac\pi2-1\right)^{\frac nx-k}\frac{d^{n-1}}{dm^{n-1}}x^{-m} dx=\sum_{m=0}^\infty\sum_{n=1}^\infty\sum_{k=0}^nS_k^{(m)}s_{n-1}^{(k)} (-1)^{k+n+1}\frac{n^m}{n!}\left(\frac\pi 2-1\right)^{-k}\frac{d^{n-1}}{dm^{n-1}}\left(b^{1-m}\E_{2-m}\left(-\frac nb\ln\left(\frac\pi2-1\right)\right)\right)\end{aligned}$$
which is supported by the two expressions being equal; much more computation is needed for more terms in the linked series, so we approximate the area using the integral of the series. Region $1,2,3$ add to $ab$, so region $2$’s area is $ab$ minus region $1$’s and $3$’s areas.
Therefore the shaded area should be:
$$\boxed{2ab-a^2-2\sum_{m=0}^\infty\sum_{n=1}^\infty\sum_{k=0}^nS_k^{(m)}s_{n-1}^{(k)} (-1)^{k+n+1}\frac{n^m}{n!}\left(\frac\pi 2-1\right)^{-k}\frac{d^{n-1}}{dm^{n-1}}\left(b^{1-m}\E_{2-m}\left(-\frac nb\ln\left(\frac\pi2-1\right)\right)\right)\approx0.00985}$$
For the rest of the area, we need a series or parametrization for the curve’s other branch. Although an approximation would be simpler, this is an exact answer.