Evaluate `int_(-1)^((3)/(2))|xsin(pix)|dx`

Explanation

`|x sin pi x|={((-x)(-sinpix),"if",-1lexlt0),(xsinpix,"if",0lexle1),(x(-sinpix),"if",1ltxle3//2):}`
`:.int_(-1)^(3//2)|x sin pi x|dx`
`=int_(-1)^(0)sin pix dx +int_(-1)^(1) x sin dx +int_(1)^(3//2)(-x sin pi x)dx`
`=int_(-1)^(1)x sin pi x dx - int_(1)^(3//2), x sin pi x dx`
`=2int_(0)^(1)x sin x dx-int_(1)^(3//2)xsin pix dx`
`=2[{x((-1)/(pi))cospix}_(0)^(1)-int_(0)^(1)1((-1)/(pi))cospix dx]`
`-{x((-1)/(pi))cospix}_(1)^(3//2)+int_(1)^(3//2)1((-1)/(pi))cosx dx`
`=(2/(pi))+(2/(pi^(2)))[sin pix]_(0)^(1)+{3/((2pi))}"cos"3/2pi+(1/(pi))`
`-1/(pi^(2))[sinpix]_(1)^(3//2)`
`=(2//pi)+0+0+(1//pi)+(1//pi^(2))`
`=(3pi+1)//pi^(2)`