Adjusting bounds of integration after a substitution with mutually dependent variables for a double integral I really have trouble with figuring out the correct bounds in such cases. Consider the substitution
$$ x = function_{1}(u, v),\\y = function_{2}(u, v) $$ for the integral $$ \int_{p}^{q} \int_{j(y)}^{e(y)}f(x, y)dxdy. $$ How am I supposed to proceed? I understand, having searched the internet, that "each case is different," but is there anything that could be thought of as a general rule? Examples I found that were supposed to explain Jacobians quite often (if not always) skip the adjustment of bounds. Of course, I could just plug in the newly introduced functions right away but then the outer integral's bounds depend on the inner integral's iterating variable, which is not supposed to happen.
Since "each case is different," I came up with a substitution the explanation of which I would appreciate. Please, do show me every detail of changing the integrals' intervals, the more detail the better. And any visuals will be much appreciated as well!
The example:
$$ \int_{a}^{b} \int_{1+b-y}^{1+b+y} \frac{1}{1-x^{2}y}dxdy,\\x = \sqrt2\frac{\tan v}{u},\\y = \frac{u^{2}\cos^{2}v}{2}. $$
 A: The general change of variables formula for double integrals is:
$$
\int_{\varphi(U)}f(x,y)dxdy=\int_Uf(\varphi(u,v))|\det(D\varphi)(u,v)|\;dudv\tag{1}
$$
where we assume nice properties of $f$, $U$ and $\varphi$ that this formula holds.
Calculating the new integrand $f(\varphi(u,v))|\det(D\varphi)(u,v)|$ under the change of variables is very straightforward.
In order to set up the integral properly, you need to figure out the corresponding $\varphi$, $U$, and $\varphi(U)$; this is the difficult step to calculate explicitly if it is possible at all in general.

Let us look at your specific example. Set $\displaystyle\varphi(u,v)=(\frac{\sqrt{2}\tan v}{u},\frac{u^2\cos^2v}{2})$. Then
$$
\varphi(U)=\{(x,y)\mid \alpha(y)\le x\le \beta(y), a\le y\le b\}, \quad
\alpha(y)=1+b-y,\quad \beta(y)=1+b+y\;.
$$
Now what you need is to find $U$ by applying the inverse of $\varphi$ to the region $\varphi(U)$.

Note that your region $\varphi(U)$ depends on the parameters $a$ and $b$. I don't see an explicit way to express $U$ except some easy observations that
$$
\sin^2v=x^2y,\quad u^2=\frac{2y}{1-x^2}
$$
which in principle tells you what $U$ is.
In general, this step is not easy if it is possible at all. This is the reason why "each case is different" and there is no one-fits-all procedure (to find $U$). If you recall, one needs special geometry of the region in order to apply the technique of polar coordinates for double integrals. Also, in order to write a double integral in terms of iterated integrals, one must have "normal domains".
For examples that one can work out $U$ explicitly, see this page and the references therein.
