changing order of int to solve Change the order of integration in the following integral: $$\int \limits_0^{2\pi} dx \int \limits_0^{\sin x} f(x,y) \, dy$$
I am very stuck on this. So from the above, we know $0<y<\sin x$ and obviously, $0<y<\sin x \leq 1$. And from this, $\arcsin y < x< \pi/2$. But am I even going the right direction?
 A: If $0\le y\le 1$ you have $x$ running from $\arcsin y$ up to $\pi - \arcsin y$.
If $-1\le y\le 0$ you have $x$ running from $\pi + \arcsin(-y)$ up to $2\pi - \arcsin(-y)$.
Just draw the picture and you'll see this. Draw the graph of the sine function.
For a value of $y$ between $0$ and $1$, there is a point on the $y$-axis.  That point is on a horizontal line above the $x$-axis, but below the highest point on the graph of the sine function.  It intersects the graph at two points.  One of those is $\arcsin y$.  The other is just as far to the left of $\pi$ as the first one is to the right of $0$, so it is $\pi-\arcsin y$.
Then do the same for the case where $-1\le y\le 0$.
Hence you have
$$
\int_0^1 \left( \int_{\arcsin y}^{\pi-\arcsin y} f(x,y)\,dx \right) \,dy - \int_{-1}^0 \left( \int_{\pi+\arcsin(-y)}^{2\pi-\arcsin(-y)} f(x,y)\,dx \right)\,dy
$$
The reason for the minus sign is that as $x$ runs from $\pi$ to $2\pi$, the integral $\displaystyle\int_0^{\sin x}$ is an integral from a larger number, $0$, to a smaller number $\sin x$.
