Proving linearity and time-invariance 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.
Your proof is correct, and the result is that the system is time varying because the response to $x(t-t_0)$ does not in general equal a delayed response to the input $x(t)$. Of course, for the special case $t_0=2\pi k$ the delayed response to $x(t)$ equals the response to $x(t-t_0)$, but for time-invariance this equality must hold for any $t_0$.