Given integral $\int \limits_0^{\pi}dx \int \limits_{0}^{2\sin x} f(x,y)dy$. Change the order of integration.
I've drawn the graph of original integral it is just one piece of $2\sin x$ function squeezed between $[0,2]$ on x-axis.
Then I'm trying to flip the axes, and I thought it's similar to rotate the graph on $90$ degree. Than I have to do some sort of manipulations to express $y$ in terms of $x$ I guess. I thought I simply can solve $2\sin x = y \Rightarrow x = \arcsin \frac{y}{2}$ and do the same thing for lower bound, but it is 0, so I can't express $x$ in terms $y$.
So now I'm confused. I don't really understand what sort of manipulations I have to perform to find the right bounds.
What about this:
$$\int_0^{2}d y \int_{\arcsin{\frac{y}{2}}}^{\pi -\arcsin {\frac{y}{2}}} f(x,y)dx$$