Running Integral of sine and cosine functions In typical signal processing course we were taught that the integral of signal $x(t)$ is given by
$$y(t) = \int_{-\infty}^{t}x(\tau) d\tau$$
How can we use this definition to evaluate the integrals of eternal sinusoids, $x(t) = \sin (t)$ or $x(t) =\cos(t)$. I mean how do you evaluate / prove that $$\int_{-\infty}^{t}\cos(\tau)d\tau = \sin(t)$$

Thanks in advance.

Edit: I raised this question because of its (ab)use in frequency modulation(FM) for single tone signal. To give details, any FM wave of single tone signal $m(t) = \cos(\omega_0 t)$ is given by 
\begin{align}
\Phi_{FM}(t) &amp; = A\cos\left(\omega_c t + K_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right)
\end{align}
It can be written as
\begin{align}
\hat{\Phi}_{FM}(t) &amp; = A\: Re\left\{\exp\left(j\omega_c t + jK_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right)\right\} 
\end{align}
or
\begin{align}
\hat{\Phi}_{FM}(t) &amp; = A\:Re\left\{\exp\left(j\omega_c t\right) \exp \left(j K_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right)\right\} 
\end{align}
In communication systems textbooks, the argument of exponential term involving cosine term is magically written as $\dfrac{K_f}{\omega_0} \sin (\omega_0 t)$ which really fascinates me. 

Later they claim this is periodic signal and hence can be represented as Fourier series expansion involving Bessel functions etc.,

As I mentioned it is really magic and am interested in knowing this magic trick. Hence this post. Sorry for being so long....

Thanks to all of you.
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]