Generay way to find Stability of system I am trying to find a general way to find system stability.I have applied these methods and struck in confusion
For eg:
If
$$y(t)= \int_{-\infty}^{t} x(\tau) \sin(4\tau) d\tau$$
then find whether the system is stable or unstable?

Process (1):
If $$y(t) = \int_{-\infty}^{t} x(\tau)\sin(4\tau) d\tau$$
then 
$$h(t) =  \int_{-\infty}^{t} \delta(\tau) \sin(4\tau) d\tau$$
$$\implies h(t) =  \int_{-\infty}^{\infty} \delta(\tau)\sin(4\tau)  u(t-\tau) d\tau$$
$$\implies h(t) = [\sin(4\tau)u(t-\tau)]|_{\tau=0}$$ 
          $$\implies h(t)=0$$.  

so system is stable as $$\int_{-\infty}^{\infty} |h(t)| dt= 0 &lt; \infty$$ ;  

Process (2):
If $$y(t)= \int_{-\infty}^{t} x(\tau)\sin(4\tau) d\tau$$
then when $x(t)=\sin(4t)$
$$y(t)= \int_{-\infty}^{t} \sin(4\tau)\sin(4\tau) d\tau  = \infty $$ 
so the system produces unbounded output $y(t)$ for bounded input $x(t)= \sin(4t)$
Hence the system is unstable 

Thus actually what will be the system stability and which process is right and which is wrong and how it is wrong?? please explain....
Your first method assumes that the system is entirely characterized by its response to the input $x(t)=\delta(t)$. This would only be the case if the system were linear and time-invariant (LTI). In that case, the input-output relation would have the form

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{1}$$

where $h(t)$ is the system's impulse response. However, the given input-output relation is not of the form $(1)$, and hence the system is not LTI. Consequently, the system's response to $x(t)=\delta(t)$ does not say much about the system's behavior in general. That's why your first method of determining the stability is not valid.

The second method is valid. You found a bounded input signal for which the output is unbounded, and in this way you've shown that the system is not BIBO-stable.