The only reasonable way to interpret "$ds$" as a functional on tangent vectors has to be that it takes a tangent vector and spits out its length, but this is not linear. So $ds$ is not a 1-form. It still seems like a nice sort of object to think about integrating. Does $ds$ fit into a larger class of gadgets generalizing differential forms? Or it there some compelling reason that I shouldn't care about $ds$?



The "ds" which appears in an integral with respect to arclength is not a 1-form. What is it?
It is **not** a 1-form, it is a *1-density*: a function that is continuous and homogeneous of degree 1 on the tangent space of the manifold. It also happens to be convex and positive in the complement of the zero section (actually, its restriction to each tangent space is a Euclidean norm). If the norm is not Euclidean, you have the arc-length element of a *Finsler metric.* The convexity is basically necessary and sufficient for the lower semi-continuity of the length functional (Busemann-Mayer theorem).


See my answer to this [question](https://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714) for more on densities.


