Behaviour of integrator at steady state
Let's take a slightly more general input signal
$$x(t)=\sin(\omega_0t)u\left(t-\frac{\theta}{\omega_0}\right),\qquad\theta\in [-\pi,\pi)\tag{1}$$
It's straightforward to show that the response of an ideal integrator to the input $(1)$ is given by
$$y(t,\theta)=\frac{\cos(\theta)-\cos(\omega_0t)}{\omega_0}\tag{2}$$
The response $y(t,\theta)$ always consists of a time-dependent component $-\cos(\omega_0t)/\omega_0$ and a DC-term $\cos(\theta)/\omega_0$, which varies between $-1/\omega_0$ and $1/\omega_0$, depending on the time when the sinusoid is switched on. Consequently, no matter how large $t$ becomes, the component that all responses to switched sinusoids have in common is always the time dependent part, but not the DC-term.
The ideal response to a sinusoid extending from $-\infty$ to $\infty$ only consists of that time-dependent component that is common to all responses to switched sinusoids. You can imagine this response as an average of $(2)$ obtained by integrating over $\theta$:
$$\tilde{y}(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}y(t,\theta)d\theta\tag{3}$$
For the some discussion on the computation of the response of an ideal integrator to a sinusoidal input, take a look at this question and its answers.