Answer
$ - \frac{1}{{9{\pi ^2}}}$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {\theta \sin 3\pi \theta } d\theta \cr
& {\text{Integrate }}\int {\theta \sin 3\pi \theta } d\theta {\text{ by parts}} \cr
& {\text{Let }}u = \theta ,{\text{ }}du = d\theta \cr
& dv = \sin 3\pi \theta d\theta ,{\text{ }}v = - \frac{1}{{3\pi }}\cos 3\pi \theta \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\theta \sin 3\pi \theta } d\theta = \theta \left( { - \frac{1}{{3\pi }}\cos 3\pi \theta } \right) - \int {\left( { - \frac{1}{{3\pi }}\cos 3\pi \theta } \right)d\theta } \cr
& {\text{Multiply and integrate}} \cr
& \int {\theta \sin 3\pi \theta } d\theta = - \frac{\theta }{{3\pi }}\cos 3\pi \theta + \frac{1}{{3\pi }}\int {\cos 3\pi \theta d\theta } \cr
& \int {\theta \sin 3\pi \theta } d\theta = - \frac{\theta }{{3\pi }}\cos 3\pi \theta + \frac{1}{{3\pi }}\left( {\frac{1}{{3\pi }}\sin 3\pi \theta } \right) + C \cr
& \int {\theta \sin 3\pi \theta } d\theta = - \frac{\theta }{{3\pi }}\cos 3\pi \theta + \frac{1}{{9{\pi ^2}}}\sin 3\pi \theta + C \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^{1/2} {\theta \sin 3\pi \theta } d\theta = \left[ { - \frac{\theta }{{3\pi }}\cos 3\pi \theta + \frac{1}{{9{\pi ^2}}}\sin 3\pi \theta } \right]_0^{1/2} \cr
& = \left[ { - \frac{{\left( {1/2} \right)}}{{3\pi }}\cos \left( {\frac{{3\pi }}{2}} \right) + \frac{1}{{9{\pi ^2}}}\sin \left( {\frac{{3\pi }}{2}} \right)} \right] - \left[ 0 \right] \cr
& {\text{Simplify}} \cr
& = \left[ { - 0 + \frac{1}{{9{\pi ^2}}}\left( { - 1} \right)} \right] \cr
& = - \frac{1}{{9{\pi ^2}}} \cr} $$
