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.
Best Answer
You can see need for the $\sin\phi$ factor by comparing the actual area on a globe with the apparent area in the Equirectangular projection. Each square of the projection represents the same change in $\theta$ and in $\phi$ and maps to a curved quadrilateral on the globe. The closer to the poles (where $\phi = 0, \pi$), the smaller the area on the globe; roughly it is proportional to $\sin\phi$, keeping in mind that the latitude measurement differs from $\phi$ by a right angle.
maps to
Equal changes in $\phi$ represent the same distance on a (spherical) globe, but equal changes in $\theta$ (longitude) represent distances proportional to $\sin\phi$, which you can prove by finding the radius of a circle of latitude that goes around around a pole (as pointed out by @Alex).