Express the integral in order dxdy and then change to dydx My region is defined by:
$$
R=\{{(x,y) \in \mathbb{R}^2: y \geq2-x^3 \wedge y\geq x^2 \wedge y\leq \sin(\tfrac{\pi}{2}x)+4 }\}
$$
I am expressing the integral as:
$$
\int\int_RfdA = \int_{-1}^1 \int_{2-x^3}^{\sin(\frac{\pi}{2}x)+4} f(x,y)dydx + \int_1^2 \int_{x^2}^{\sin(\frac{\pi}{2}x)+
4}f(x,y)dydx.
$$
And changing the order:
$$
\int_1^3\int_{\sqrt[3]{2-y}}^{\sqrt{y}}f(x,y)dxdy + \int_3^5\int_{\frac{2}{\pi}\arcsin(y-4)}^{\sqrt{y}}f(x,y)dxdy.
$$
I'm not sure if I'm thinking about it.

Your integral set-up in order $dy \ dx$ is correct. In the order $dx \ dy$, there are two issues with your set-up you need to correct.

i) If you look at the graph, for $4 \leq y \leq 5$, $x$ is bound below and above by sine curve and not between sine curve and the parabola. So you need to break $3 \leq y \leq 5$ into two.
ii) For $1 \leq y \leq 3$, you have lower bound of $x$ as $ \sqrt[3]{2-y}$. $x$ is positive for $1 \leq y \leq 2$ and negative for $2 \leq y \leq 3$. As the lower bound is within roots, you should split it.
So here is how I would set up,
$ \displaystyle \int_1^2 \int_{(2-y)^{1/3}}^{\sqrt y} f(x, y) \ dx \ dy \ + \  \int_2^3 \int_{-(y-2)^{1/3}}^{\sqrt y} f(x, y) \ dx \ dy$
$ + \ \displaystyle \int_3^4 \int_{\frac{2}{\pi} \arcsin(y-4)}^{\sqrt y} f(x, y) \ dx \ dy$
$+ \ \displaystyle \int_4^5 \int_{\frac{2}{\pi} \arcsin(y-4)}^{\frac{2}{\pi} (\pi - \arcsin(y-4))} f(x, y) \ dx \ dy$
