they do cancel it out the 2cosu and because 4sin^2(u) = 2(1-cos2u), integrating w.r.t.u you get 2u - sin2u = 2u - 2sinucosu
Also, because its an indefinite integral, you gotta resubstitute back in terms of x. so remembering your substitution x = 2sinu :. u = sin^-1(x/2)