Concerning setting the precision of c: Have you seen Rationalize and SetPrecision? Consider using c = Rationalize[1.2], c = Rationalize[1.2, 0], c = Rationalize[Rationalize[1.2], 0] or c = SetPrecision[1.2, Infinity]. And so forth. You can substitute for 1.2 the computed value/variable in your actual use-cases.
I get an accuracy of 199.5 correct digits for c = 1.2`200 and 199.9 digits for c = 1.2`201 in V11.1.1. For WorkingPrecision -> 200, the default PrecisionGoal for a one-dimensional integral is half (I think), or 100 digits. In both cases, I'm getting considerably more accuracy than requested. So I can't really comment on your results, since I cannot reproduce them. -- Aha, I may have a guess. The 0.+<<18>>... in the error message has a machine-precision float 0., which suggests a machine-precision parameter was used. There are none in your code, but if I set n = 7. (to a machine-precision float, instead of an exact integer 7), then the following reproduces the expression in your error message.
Sqrt[c^2 + k^2] ChebyshevU[7., k/(I c)]^2 // Short[#, 3] &
Note the differences between WorkingPrecision, PrecisionGoal, and AccuracyGoal in NIntegrate. With PrecisionGoal -> p and AccuracyGoal -> a, NIntegrate tries to make the estimated error for a result
$I$ less than
$10^{-a} + |I|\,10^{-p}$. They control the algorithm. WorkingPrecision on the other hand controls the rounding error in the computations. For convenience, when PrecisionGoal and AccuracyGoal are set to Automatic, Mathematica makes them depend on WorkingPrecision, usually setting them to half the setting for WorkingPrecision. One generally needs a higher working precision than the precision goal.
