Say we have a positive point charge located somewhere. Then the electric field due to this point charge is $\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r} $ , now considering potential is $V = -\int_{a}^{b}\mathbf{E}\cdot d\mathbf{s}$ then the potential at an arbitrary point $r_1$ away from the charge(considering $V_\infty=0$) is $$V = -\int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r} \cdot d\mathbf{s}$$. Now considering we are going from infinity to $r_1$, $d\mathbf{s}$ should be pointed radially inward, while the electric field should be pointed radially outward giving a $cos(180)$ so I get

$$V = -\int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}cos(180) dr$$

$$= \int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} dr$$ which ends up being $$V=-\dfrac{Q}{4\pi \epsilon_0r_1}$$.

What am I missing here?