multivariable calculus - How to change the order of integration for this?

So I can change order of integration for simple functions through the use of diagram but how do i do it for

$$\int_{0}^{\pi}\int_{0}^{\sin x}f(x, y)dydx?$$

So y goes from 0 to 1 but the functions needs to be split at $\pi/2$ when we consider the $x$ direction so in the $x$ direction it goes from $x=\sin^{-1}y$ to $\pi/2$ and then $\pi/2$ to $x=\sin^{-1}y$ or something? This question has been asked before but i don't think the asker does it the way i do so didn't see a good answer

Answer

Here, in red is the graph of

$$y=\pi-\arcsin(x)$$

To apply this to the problem, we need the inverse

$$x=\pi -\arcsin(y)$$

which is 'inverse' of $y=\sin(x)$ for $\frac{\pi}{2}\le x\le\frac{3\pi}{2}$

This is the inverse function you need to integrate the other half when reversing the order of integration.

Addendum: Since you are still unsatisfied with the answers given so far, I will add the following additional bit of explanation.

Clearly, the sine function is not one-to-one on the interval $\left[0,\pi\right]$. However, it is one-to-one on the intervals $\left[0,\frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2},\pi\right]$ and clearly the inverse functions on those two intervals will involve $\sin^{-1}x$.

Since the inverse function is defined only on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ we know that $X=\sin^{-1}$. But for $x$ in $\left[\frac{\pi}{2},\pi\right]$, this cannot be the case.

Now notice that each $x$ in the interval $\left[0,\frac{\pi}{2}\right]$ has a "mirror image" $X$ in the vertical line $x=\frac{\pi}{2}$.

Since $\frac{\pi}{2}$ is half-way between $X$ and $x$ it is their average. So

$$\begin{eqnarray} \frac{X+x}{2}&=&\frac{\pi}{2}\\ X+x&=&\pi\\ x&=&\pi-X\\ x&=&\pi-\sin^{-1}y \end{eqnarray}$$

So when reversing the order of integration we know that $y$ moves between the values of $0$ and $1$ and $x$ goes between the values of $\sin^{-1}y$ and $\pi-\sin^{-1}y$, giving

$$\int_0^1\int_{\sin^{-1}y}^{\pi-\sin^{-1}y}f(x,y)\,dxdy$$