This indefinite integral is simple if we note that `sin(6x) dx = -1/6 d(cos(6x)).` Formally, perform the variable substitution `u = cos(6x),` then `du = -6sin(6x)` and the integral becomes
`int e^u (-1/6) du = -1/6 e^u + C = -1/6 e^cos(6x) + C,`
where `C` is an arbitrary constant.
You haven't shown the limits of integration, but the function under integral is continuous and bounded on the whole axis, therefore this integral is not improper at any finite interval.
If we consider it at any infinite interval (to `+oo` or to `-oo,` or both), then it diverges because the expression `-1/6 e^cos(6A)` is periodic and has no limit when `A -> oo.`