The relations of the value of the constant of integration is zero between a DC component
$$i(t)=v_0/(\omega L) \times (-cos(\omega t) + C)$$
Current cannot change instantly for an AC sinusoidal voltage supply ,so we can use $$i(t)=0 $$at t=0
And we get $$C=1$$ and hence total response would be
$$i(t)=\frac{v_0}{L}\frac{(-\cos(\omega t)+1)}{\omega}$$
$$\frac{v_0}{L\omega}$$ is DC component of the current .
Why we ingnore   (DC component)at  steady state analysis ?
for a pure inductive circuit we cannot ignore dc component even for steady state analysis but practically pure inductive circuit is impossible and small resistor is always present so the constant of integration for R-L circuit is $$A\exp^{(-R/L)t}$$
And at steady state it decays to zero ,and that why for steady state analysis of pure inductive circuit constant of integration is assumed to be zero