How do I reverse the order of integration for this double integral? $$\int_0^{\pi/2}\int_{\sin(x)}^2dydx$$ This is the integral in question. My closest was $$\int_0^{2}\int^{\sin^{-1}(y)}_2dxdy$$
but I believe this is wrong.
Sorry for the bad formatting, It's my first time using it properly
As saulspatz pinted out in the comment, you will have to split the integral into two if you switch the order:
$$
\int_0^1\int_0^{\sin^{-1}(y)}dxdy+\int_{1}^2\int_0^{\pi/2}dxdy
$$
Actually,
$$\int_0^{\pi/2}\int_{\sin(x)}^2\,dy\,dx = \int_0^{\pi/2}\left(2-\sin(x)\right)\,dx = \pi-1$$