Set \(y=\pi x\), so that \(\dfrac{\mathrm dy}\pi=\mathrm dx\) and the integral is equivalent to
\(\displaystyle\int_0^{1/2}x\cos\pi x\,\mathrm dx=\frac1{\pi^2}\int_0^{\pi/2}y\cos y\,\mathrm dy\)
Now integrate by parts, setting
\(u=y\implies\mathrm du=\mathrm dy\)
\(\mathrm dv=\cos y\implies v=\sin y\)
So you have
\(\displaystyle\frac1{\pi^2}\int_0^{\pi/2}y\cos y\,\mathrm dy=\frac1{\pi^2}y\sin y\bigg|_{y=0}^{y=\pi/2}-\frac1{\pi^2}\int_0^{\pi/2}\sin y\,\mathrm dy\)
\(=\displaystyle\frac{\frac\pi2\sin\frac\pi2-0}{\pi^2}+\frac1{\pi^2}\cos y\bigg|_{y=0}^{y=\pi/2}\)
\(=\displaystyle\frac1{2\pi}+\frac{\cos\frac\pi2-\cos0}{\pi^2}\)
\(=\displaystyle\frac1{2\pi}-\frac1{\pi^2}\)
