Knowing that $\int_0^\pi\frac{x\sin^{2022}x}{\sin^{2022}x+\cos^{2022}x}\, \mathrm dx=\frac{\pi^a}{b} \ (a, b \in \mathbb N)$, calculate $2a^2 + 3b^3$. 
Knowing that $\displaystyle \int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \dfrac{\pi^a}{b}$, where $a$ and $b$ are positive integers, calculate the value of $2a^2 + 3b^3$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Let's pretend I know what's going on.
As $f(0) = f(\pi)$ where $f(x) = \dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}$, it can be proven that $$\displaystyle \int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \dfrac{\pi}{2}\int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx$$
You know what's the deal with years being used in problems? They're interchangeable.
Consider sequence $(u_n)$ where $\displaystyle u_n = \int_0^\pi\dfrac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}\, \mathrm dx, \forall x \in \mathbb N$.
(Why shouldn't it be $\displaystyle u_n = \int_0^\pi\dfrac{\sin^{n}x}{\sin^{n}x + \cos^{n}x}\, \mathrm dx, \forall x \in \mathbb N$? Well, $x = -\dfrac{\pi}{4} + k\pi, k \in \mathbb Z$ is a root for all equations $\sin^{2n + 1}x + \cos^{2n + 1}x = 0, n \in \mathbb N$, which means the integral does not converge at $x = \dfrac{3\pi}{4}$.)
We have that $$\begin{aligned} u_{n + 1} - u_n &= \int_0^\pi\dfrac{\sin^{2n + 2}x}{\sin^{2n + 2}x + \cos^{2n + 2}x}\, \mathrm dx - \int_0^\pi\dfrac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}\, \mathrm dx\\ &= \int_0^\pi\dfrac{\sin^{2n}2x(2\sin^2x - 1)\, \mathrm dx}{2^n\sin^2x(\sin^{2n + 2}x + \cos^{2n + 2}x)(\sin^{2n}x + \cos^{2n}x)}, \forall n \in \mathbb N \end{aligned}$$...I think.
There's also $$\displaystyle \int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx + \displaystyle \int_0^\pi\dfrac{\cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx = \displaystyle \int_0^\pi\, \mathrm dx = \pi$$
Perhaps a step in the right direction would be solving $\displaystyle \int_0^\pi\dfrac{\sin^{2022}x - \cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \mathrm dx$.
It's also interesting to note that $\displaystyle \int\dfrac{\sin x - \cos x}{(\sin x + \cos x)^2}\, \mathrm dx = \dfrac{1}{\sin x + \cos x}$ and $\displaystyle \int\dfrac{2}{(\sin x + \cos x)^2} = \dfrac{\sin x - \cos x}{\sin x + \cos x}$.
But that's all for now, this was a mess. As always, thanks for reading, (and even more so if you could help), and have a great tomorrow~
By the way, the choices were $32, 194, 200$ and $100$. WolframAlpha gives the result of $\dfrac{\pi^2}{4}$, so the answer is $2 \times 2^2 + 3 \times 4^3 = 200$.

Continue with
\begin{align}
&\int_0^\pi\dfrac{x\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\,dx 
= \dfrac{\pi}{2}\int_0^\pi\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx\\
 = &\ \dfrac{\pi}{2}\int_0^\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx
 + \frac{\pi}{2}\int^\pi_\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, \overset{x\to \frac\pi2+x}{dx}\\
 = &\ \dfrac{\pi}{2}\int_0^\frac\pi2\dfrac{\sin^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx
+\frac{\pi}{2}\int_0^\frac\pi2\dfrac{\cos^{2022}x}{\sin^{2022}x + \cos^{2022}x}\, dx\\
= &\>\frac\pi2\int_0^\frac\pi2 dx=\frac{\pi^2}4
\end{align}
