where does the 2 go?
But ds and dx are just markers reflecting what variable we using in the intergrand. So if the variable we switch it to is twice that of the one we switched it from, I don't see how that effects anything in the integrand.
OK, so the way I'm thinking about it now is that when you do the substitution you're changing the integrand into an expression that operates on a variable that is twice that of the original variable. So to keep the integrand equal to the value of the original expression you need to keep the balance somehow. When I put the above integrand into wolfram, but with a 1 in place of the 2, it pulls out a 1/2 to do this. So there is a link here obviously between the s being twice that of u, but I can't seem to quite grasp it. Does simply dividing the whole integrand by 2 keep the integrand equal to the first variable, it obviously does, but I can't quite grasp why is should. Can anyone explain it to me?