When you integrate in spherical coordinates, the differential element isn't just $ \mathrm d\theta \,\mathrm d\phi $. No. It's $\sin\theta \,\mathrm d\theta \,\mathrm d\phi$, where $\theta$ is the inclination angle and $\phi$ is the azimuthal angle.

For example, attempting to integrate the unit sphere without the $\sin\theta$ term:

$$
\int\_0^{2\pi}\int\_0^{\pi} \mathrm d\theta \,\mathrm d\phi = 2 \pi^2.
$$

With the $\sin\theta$ term you get

$$
\int\_0^{2\pi}\int\_0^{\pi} \sin\theta \,\mathrm d\theta \,\mathrm d\phi = 4 \pi.
$$

But I'm puzzled why you need to multiply the differential solid angle by $\sin\theta$. It would seem it's because the chunks at the north/south pole are "worth less" than the chunks at the equator. That kind of makes sense because they will be closer together.