Let $D$ be the region in the first quadrant of the $x,y$-plane, bounded by the curves $y=x^2$, $y=x^2+1$, $x+y=1$, and $x+y=2$.
I want to find a change of variables for $\iint_D x\sin(y-x^2) \, dA$.
Since $y-x^2=0$, $y-x^2=1$, my inclination is to let $u=y-x^2$, and since $x+y=1$ and $x+y=2$ I want to let $v=x+y$. However, when solving for $x$ in terms of $u$ and $v$ I get a horrid looking thing.
Could someone provide me with a better way of approaching this problem? Thank you!
Here's half an answer. Maybe you went down this path already:
If you let $u=x^2, v=x+y$ some things are less horrid. First, the Jacobian is $\frac{1}{2x}$ which cancels the first $x$ in the integrand. The pre-image $D^*$ of the domain is bounded by $v=1, v=2, v=u+\sqrt{u},$ and $v=u+\sqrt{u}+1.$
At first glance $D^*$ tempts you to integrate in the order $du \; dv.$ But you run into the same horridness solving those equations for $u$. But happily, the 4 points of intersections are not that bad and (here's the good part) the middle two have the same $u$ coordinate $a=(3-\sqrt{5})/2$. So you can break $D^*$ into two regions.
First region is $u=0$ to $a$, $v=1$ to $u+\sqrt{u} +1$.
Second region is $u = a$ to $1$, $v=u+\sqrt{u}$ to $2$.
Everything works swimmingly until the end when there are two integrals:
$$\int_0^a \frac{1}{2} \cos(1-\sqrt{u}-u) \; du$$
and
$$\int_a^1 \frac{1}{2} \cos(2-\sqrt{u}-u) \; du.$$
I was hoping for a clever substitution that would change one of them into the other and then they'd cancel out. I've had no luck with that so far.
I have added an answer to an other question which solves this question as well or shows where the problem is in general for this integral to be evaluated: https://math.stackexchange.com/a/2876108/541516
An other interesting variable change which can take us much further and show that this integral cannot be evaluated to closed form solution without series expansion, would be to remove away $x$ in the integral. See that we have $x^2$ inside $\sin(\cdot)$ which can make this happen.
Let, $k = x^2$ and keeping $y$ as it is, which gives a jacobian of $1/2x$ and cancels with the $x$. Now, although we have $\sin(y-k)$ kind of structure which we can expand by trigonometric identity of $\sin(A-B)$ into two parts when function of $k$ and $y$ are independent, it is better to do an other variable change which makes integral limits associated with $y$ to fixed limits.
See that now the limit curves are of the form $y-k = c$ and $\sqrt k + y = c$.
So, we can take new variable $t = y-k$ which actually does this, making the limits for $t$ to be 0 to 1. Limits for $k$ are taken to be dependent on t and of the form $\sqrt k + k + t = c$. Now the inner function is just $sin(t)$ which can be pulled out of the integral with $k$. Now to evaluate the integral with $k$ we just have to evaluate the limits i.e. $\sqrt k + k + t = 1$ and $\sqrt k + k + t = 2$. If we solve the quadratic equations in $\sqrt k$ and then take square the positive of the quadratic solution which is ($-1 + \sqrt{1 + 4(c-t)}$)/2, we get, constants, $t$ terms and $\sqrt{c-t}$ kind of terms. And remember a $sin(t)$ is waiting for us and limits are 0 to 1 in variable t!
Now, $\sin(t)$ can be evaluated, $t \sin(t)$ is well known by integration by parts. The problematic part is the terms with the structure $\sqrt{c-t} \sin(t)$. To get rigid of c (which is 1 and 2 here), we can do variable transformation and $p = c-t$ which brings us a structure $\sqrt p \sin(c - p)$ which similar to the general structure, $\sqrt p \sin(p)$ through expansion of $\sin(c - p)$.
So, now to this $\sqrt{p} \sin(p)$ when we do integration by parts, this ends up with having an integral of $cos(p)/\sqrt{p}$ to be evaluated, which on variable transformation $p = u^2$ gives integral of $\cos(u^2)$ which further more cannot be simplified by hand and is usually termed Fresnel integral which has series expansion ( see https://en.wikipedia.org/wiki/Fresnel_integral).
PS: I have focused only on the hard terms and ignored signs, constants and other easy terms coming from integration by parts just for transparent idea of what creates problem, hope it is understandable, let me know if some part is unclear.