Behaviour of integrator at steady state I wanted to calculate response of integrator  of sinusoidal input at steady state via these two methods as mention in image  but these two methods give two different answers at steady state, so where I'm going wrong? What concept I'm missing?
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.