Evaluating $\iint_D\frac{\sin y}{y}dxdy$, where $D$ is the region enclosed by $y=2$, $y=1$, $y=x$, $y=2x$ I have been given this exercise: Calculate the double integral:

$$\iint_D\frac{\sin(y)}{y}dxdy$$
Where $D$ is the area enclosed by the lines: $y=2$, $y=1$, $y=x$, $2y=x$ (not $y = 2x$).

Visualising $D$ is easy. You can split D in two sub areas and get the bounds for the integrals. The problem I face is:
Let's split D in two sub areas, $D_1$ and $D_2$. $D_1$ is the left, upright triangle of $D$ and $D_2$ is the right, upside down one.
Then $D_1$ is defined by the lines $y=1$, $y=x$, and $x=2$.
You can express the area in a $y$-normal form as:
$$\begin{align}
  1 \le y \le 2\\
  y \le x \le 2
\end{align}$$
then the integral can be written as
$$ \begin{align}
  &\int_1^2\int_y^2\frac{\sin(y)}{y}dxdy \\
  &=\int_1^2\frac{\sin(y)}{y}[x]^2_y \space dxdy \\
  &=\int_1^2\frac{2\sin(y)}{y} - \sin(y)dy \\
  &=2\int_1^2\frac{\sin(y)}{y}dy -\int_1^2 \sin(y)dy \\
\end{align}$$
The second integral is trivial, but in the first one is not. I have tried substituting, integrating by parts but to no avail. What am I doing wrong?
Any answer is really appreciated.

Note that the integrand does not depend on $x$:
$$
\int\int_D \frac{\sin y}{y} dx dy
= \int_1^2   \left( \int_{y/2}^y dx\right) \frac{\sin y}{y} dy
= \int_1^2  \frac{y}{2}\frac{\sin y}{y}  dy
$$
$$
= \int_1^2 \frac{\sin y}{2}  dy = \frac{\cos 1 - \cos 2}{2}
$$
