Evaluate `intsinxsin2xdx`.

Explanation

`intsin2xsin3xdx=(1)/(2)int2sin3xsin2xdx`
`=(1)/(2)int[cos(3x-2x)-cos(3x+2)]dx`
` [ :' 2sinA sin B=cos(A-B)-cos(A+B)]`
`=(1)/(2)int[cosx-cos5x]dx`
`=(1)/(2)intcosxdx-(1)/(2)intcos5xdx`
`=(1)/(2)sinx-(1)/(2)(sin5x)/(5)+C`
`=(sinx)/(2)-(sin5x)/(10)+C`