How rough can differential form, manifold and chain be for Stokes theorem to hold? When I first learned Stokes theorem, everything is assumed to be smooth to prevent any strange things happen. But to apply to more cases, I may need to use a version of Stokes theorem that holds for rougher forms, chains and manifolds. For instance, when I learned Cauchy integral theorem, the paths and the analytic functions are only assumed to be $C^1$.
Does Stokes theorem hold for merely $C^1$ forms, chains and manifolds? I think so because we only exterior differentiate once in the equation of Stokes theorem, and the pullback by the chain (parametrisation of surfaces) only uses the first derivative, while the continuity of first derivative is added to ensure the exterior derivative ($d\omega$) and the pullback form ($(\partial c)^*\omega$) is integrable.
Edit: I see that Stokes theorem holds for manifolds with corners. I suppose these are equivalent to piecewise smooth surfaces, lines, etc. However, it is not all of what I am asking about. I am asking whether Stokes theorem holds when the differential form, manifold and chain are simply assumed to be $C^1$.

In Lang's Real Analysis you'll find Stokes's Theorem stated for a $C^2$ manifold and $C^1$ form $\omega$. He goes on to discuss what to do with singularities. 
Although Lang asserts existence of a $C^2$ partition of unity using the $C^2$ hypothesis, a $C^1$ partition of unity suffices. However, as @MatheinBoulomenos pointed out to me in chat, you need a $C^2$ chart in order to be sure that the pullback of the $C^1$ form (bumped off by the $C^1$ partition of unity) is still a $C^1$ form. (Lang is very sloppy about paying attention to details in this regard.) Then it's the usual computation.
