How to you solve this indefinite integral from 0 to 1? ∫ (xsin (πx^2)) dx

This is another view of substitution ruleu=πx^2∫ (xsin (πx^2)) dx = ∫ (xsin (u)) dx = ∫ (xsin (u)) dx [du / du] = ∫ (xsin (u)) du [ 1 / (du/dx) ] = ∫ (xsin (u)) du [ 1 / 2πx ] = (1/2π) ∫sin (u) du = (1/2π) cos (u) = (1/2π) cos (πx^2)

notice if you diff term u (πx^2) and the result can eliminate (divide) extra term (here is x) then this rule is applicable.