newtonian gravity - rotational oblateness

I am trying to compute the amount of oblateness that is caused by planetary rotation. I picture the force of gravity added to the centrifugal force caused by the rotation of the planet as follows:

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That is, at the point in question, at latitude $\phi$, the distance from the axis of rotation is $r\cos(\phi)$. Thus, the centrifugal force would be $\omega^2r\cos(\phi)$ in a direction perpendicular to the axis of rotation. The radial and tangential components would be $\omega^2r\cos^2(\phi)$ and $\omega^2r\cos(\phi)\sin(\phi)$, respectively.

My assumption is that the surface of the planet would adjust so that it would be perpendicular to the effective $g$; that is, the sum of the gravitational and centrifugal forces. This would lead to the equation

$$\frac{\mathrm{d}r}{r\,\mathrm{d}\phi}=-\frac{\omega^2r\cos(\phi)\sin(\phi)}{g-\omega^2r\cos^2(\phi)}$$ We can make several assumptions here, and I will assume that $\omega^2r$ is small compared to $g$. Thus, we get $$\int_{\text{eq}}^{\text{np}}\frac{\mathrm{d}r}{r^2} =-\frac{\omega^2}{g}\int_0^{\pi/2}\cos(\phi)\sin(\phi)\,\mathrm{d}\phi$$ which leads to $$\frac1{r_{\text{np}}}-\frac1{r_{\text{eq}}} =\frac{\omega^2}{2g}$$ and $$1-\frac{r_{\text{np}}}{r_{\text{eq}}} =\frac{\omega^2r_{\text{np}}}{2g}$$ However, numerical evaluation and Wikipedia seem to indicate that this should be twice what I am getting. That is, $$1-\frac{r_{\text{np}}}{r_{\text{eq}}} =\frac{\omega^2r^3}{Gm} =\frac{\omega^2r}{g}$$ What am I doing wrong?