Fourier series of a square pulse (simple simplification)
You could rewrite $x(t)$ as a rectangular function before doing the Fourier transform. You could deduce the width and center of the rect signal directly from the given expression but it would be a easier to just draw the signal and deduce the values from the graph.

Your rect signal will be in the following form:

$x(t) = rect(\frac{t-t_o}{W})$ 

where:

$t_0$ - center of the rectangular signal 

W - width(duration) of the pulse

From this point you would use a Fourier transform table and some Fourier transform theorems (involving scaling and delay) to get the $sinc$ version of the $X(\omega)$.

Or, if you prefer, you could just modify a little bit the expression you already have to get the expression where you could use Euler's formula as Dilip Sarwate commented.

$X(\omega)=1 \cdot \frac{1}{j\omega}(e^{-j\omega} - e^{-j2\omega})= e^{-j1.5\omega}e^{+j1.5\omega} \frac{1}{j\omega}(e^{-j\omega} - e^{-j2\omega}) = e^{-j1.5\omega}\frac{1}{j\omega}(e^{+j0.5\omega} - e^{-j0.5\omega})$

From this point it should be easy to get the result.