Running Integral of sine and cosine functions
I would not teach integrals this way. 

Issues might arise form the definition of terms. In standard calculus, a (note: "a", not "the) primitive function (also called antiderivative or indefinite integral) of a continuous function $f$ is a kind of converse of the concept of derivation. SO: 


  If $F$ is "a" differentiable function  whose derivative is equal to
  $f$, then $F$ is a primitive (function) of $f$. There exists many such
  functions, up to a constant $c$, because: $(F(t)+c)' = (F(t))' =f(t)$.


Here, you can check that $t\mapsto \sin(t)+ c$ is differentiable, with derivative $t\mapsto \cos(t)$. 

Now (if I remember well), suppose that $f(x)$ is a continuous function, AND that it is  "integrable" over $\mathbb{R}$, then $F(t) = \int_{-\infty}^t f(\tau)d\tau\;$ exists, and is a differentiable function. THEN, it is also a primitive of $f(t)$ (among others).

But here, it is not possible, namely because $t\mapsto \cos(t)$ is not "integrable" over $\mathbb{R}$, and thus the integral expression makes no sense.

I would say that one should not define "integrals" the way you wrote it, if one wants mathematical correctness. However, for causal signals, in a lousy engineer perspective, it may be intuitive. But the cosine is not causal.

[Note: there are more general definitions of primitives]