Answer
$$ - \frac{1}{3}\cos \frac{3}{2}x - \cos \frac{1}{2}x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin x\cos \left( {x/2} \right)} dx \cr
& {\text{product }}\sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& \sin x\cos \left( {x/2} \right) = \frac{1}{2}\left( {\sin \left( {\left( {1 - \frac{1}{2}} \right)x} \right) + \sin \left( {\left( {1 + \frac{1}{2}} \right)x} \right)} \right) \cr
& \sin x\cos \left( {x/2} \right) = \frac{1}{2}\left( { - \sin \left( {\frac{1}{2}x} \right) + \sin \left( {\frac{3}{2}x} \right)} \right) \cr
& \int {\sin x\cos \left( {x/2} \right)} dx = \frac{1}{2}\int {\left( {\sin \frac{3}{2}x - \sin \frac{1}{2}x} \right)} dx \cr
& {\text{sum rule}} \cr
& = \frac{1}{2}\int {\sin \frac{3}{2}x} dx - \frac{1}{2}\int {\sin \frac{1}{2}x} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\left( { - \frac{2}{3}\cos \frac{3}{2}x} \right) - \frac{1}{2}\left( {2\cos \frac{1}{2}x} \right) + C \cr
& = - \frac{1}{3}\cos \frac{3}{2}x - \cos \frac{1}{2}x + C \cr} $$
