How we can obtain the time-integration of a sine signal?
No, this improper integral doesn't have a value. Since for the indefinite integral we have

$$\int\sin(\omega_0t)dt=-\frac{1}{\omega_0}\cos(\omega_0t)+C\tag{1}$$

the given definite integral would be

$$\int_{-\infty}^{t}\sin(\omega_0\tau)d\tau=-\frac{1}{\omega_0}\cos(\omega_0t)+\lim_{\tau\rightarrow -\infty}\frac{1}{\omega_0}\cos(\omega_0\tau)\tag{2}$$

but the limit in $(2)$ doesn't exist.