This question was asked in one of our quiz. Question is
Determine if the system with the input-output relationship given by
$$y(t) = \int_{t-10}^{t} \cos(\tau)x(\tau)d\tau$$
is linear and time invariant.
I was able to prove linearity but was not able to prove the time invariance of the system.
My Try
For invariance we need to show output resulted from $x(t-t_o)$ is same as $y(t-t_o)$. Let $$y_1(t) = \int^{t}_{t-10} \cos(\tau)x(\tau-t_o)d\tau.$$ Now when I change of variable to $z=\tau - t_o$, $dz = d\tau$ it leads to $$y_1(t) = \int^{t-t_o}_{t-10-t_o} \cos(z+t_o)x(z)dz$$ where I get stuck (using $\cos(A+B)$ identity doesn't seem to work well) and not able to proceed further.
For $y(t-t_o)$ integral becomes $$y_2(t-t_o) = \int^{t-t_o}_{t-10-t_o} \cos(\tau)x(\tau)d\tau$$
From here it seems to time variant system. But when $t_o=2\pi n$ it becomes time invariant. But system cannot be time invariant and time variant. That's the doubt I have with this question whether the system is time invariant or time variant and why.