Let I = \(\int\limits_0^{3/2}\)|x cos π x|dx

\(\because\) |x cos x| = \(\begin{cases}xcos\pi x&;0\leq x\leq\frac12\\-xcos \pi x&;\frac12\leq x\leq \frac32\end{cases}\)

\(\therefore\int\limits_0^{3/2}\)|x cos π x| dx = \(\int\limits_0^{1/2}xcos\pi x-\int\limits_{1/2}^{3/2}xcos\pi xdx\) 

\(=[\frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2}]_0^{1/2}\)\(-[\frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2}]_{1/2}^{3/2}\)

\((\because\int xcos\pi x= \frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2})\) 

\(=\frac1{2\pi}sin\frac{\pi}2+\frac1{\pi^2}cos\frac{\pi}2-\frac1{\pi^2}cos0\) \(-(\frac3{2\pi}sin\frac{3\pi}2+\frac1{\pi^2}cos\frac{3\pi}2-\frac1{2\pi}sin\frac{\pi}2-\frac1{\pi^2}cos\frac{\pi}2)\)

\(=\frac1{2\pi}-\frac1{\pi^2}-(-\frac{3}{2\pi}-\frac1{2\pi})\)

(\(\because\) sin \(\frac{\pi}2=1,\) sin \(\frac{3\pi}2=-1\), cos 0 = 1 and cos \(\frac{\pi}2=cos\frac{3\pi}2=0\))

\(=\frac1{2\pi}-\frac{1}{\pi^2}+\frac4{2\pi}\) 

\(=\frac{5}{2\pi}-\frac1{\pi^2}=\frac{5\pi-2}{2\pi^2}\)