Evaluate $\iint_D x \sin (y -x^2) \,dA$. 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 $x+y=2$. Using an appropriate change of variables, compute the integral
$$\iint_D x \sin (y -x^2) \,dA.$$
I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!

I'm not sure Kishores answer is precisely the same, but I try to be more illustrative and short. First of all we transform the integral by defining \begin{align}
u &= y-x^2 \\ 
v &= x^2
\end{align}
which leads to
$$
\int_{\rm D} x\sin(y-x^2) \, {\rm d}x \, {\rm d}y
=\frac{1}{2}\int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v \, .
$$
In the $(x,y)$-plane the area of integration looks like this

which in the $(u,v)$-plane becomes

so that the $v$ integral can be done trivially
\begin{align}
\frac{1}{2} \int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v &= \frac{1}{2} \int_0^1 {\rm d}u \sin u \int_{\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u}^{\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u} {\rm d}v \\
&=\frac{1}{2} \int_0^1 {\rm d}u \sin u \left\{1+\frac{\sqrt{5-4u}}{2} - \frac{\sqrt{9-4u}}{2} \right\} \, .
\end{align}
Integrals of this form lead to Fresnel-Functions
\begin{align}
&\int  \sin u \, \sqrt{1-au} \, {\rm d}u \\ = &\sqrt {\frac{\pi \,a}{2}} \left\{ \cos \left( \frac{1}{a} \right) \, {\rm C} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) + \sin \left( \frac{1}{a} \right) \, {\rm S} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) \right\} -\cos (u) \, \sqrt {1-au}
\end{align}
where $0\leq a \leq 1$. The numerical value of the integral is $\approx 0.1141123$.
In the $(u,v)$-plane
\begin{align}
&{\rm 1)} \quad 2-x \text{ is mapped on } v=\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u \\
&{\rm 2)} \quad 1-x \text{ is mapped on } v=\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u \\
&{\rm 3)} \quad x^2 \text{ is mapped on }  u=0 \\
&{\rm 4)} \quad x^2+1 \text{ is mapped on } u=1 \,  .
\end{align}
