Evaluate $\iint_D\frac{\sin x}{x}\;\mathrm dx\mathrm dy$ Evaluate, where $D=\{(x,y): 0\le x \le 1 , 0\le y \le x\}$,
  $$\iint_D\frac{\sin x}{x}\;\mathrm dx\mathrm dy$$ 
My solution and question
$$\iint_D\frac{\sin x}{x}\;\mathrm dx\mathrm dy = \int_0^1\frac{\sin x}{x}\;\mathrm dx\int_0^x\mathrm dy =\\= \int_0^1\sin x \;\mathrm dx  = [-\cos x ]_0^1=\ 1-\cos 1$$
Is it correct? Thanks in advance.

Your result is correct (and concise).  However, if you want a different route, then you can do the following.  The sine integral $\text{Si}$ is defined as follows:
$$\text{Si}(z):=\int_0^z\,\frac{\sin(t)}{t}\,\text{d}t\text{ for all }z\in\mathbb{C}\,.$$
Integration by parts gives
$$\int\,\text{Si}(t)\,\text{d}t=t\,\text{Si}(t)-\int\,\sin(t)\,\text{d}t=t\,\text{Si}(t)+\cos(t)+\text{constant}\,.$$
That is,
$$I:=\underset{D}{\int\int}\,\frac{\sin(x)}{x}\,\text{d}x\,\text{d}y=\int_0^1\,\int_y^1\,\frac{\sin(x)}{x}\,\text{d}x\,\text{d}y=\int_0^1\,\big(\text{Si}(1)-\text{Si}(y)\big)\,\text{d}y\,.$$
Consequently,
$$I=\text{Si}(1)-\int_0^1\,\text{Si}(y)\,\text{d}y=\text{Si}(1)-\Big(\text{Si}(1)+\cos(1)-\cos(0)\Big)=1-\cos(1)\,.$$
