This is a broad question which might not have a clear answer. I am aware that the incompressible NS equations can be used to approximate the compressible when the Mach number is low, and that this approximation is validated by results in functional analysis.
However, I am currently wondering what the 'use' is for this approximation. I remember during my bachelor's degree in physics lessons that we were taught that this approximation was useful as the incompressible version was 'easier' for computing (by hand) exact solutions in specific cases like pipe flow. This just seemed intuitive to me at the time, as there were of course fewer terms.
Does this hold true when computing numerical solutions though? Is the incompressible version easier in most or even any cases?
I ask as I am aware that the incompressible version can have its own set of problems (for example one can't use the barotropic assumption that the pressure is just the density to the power of some constant in the incompressible case). So perhaps no broad comparisons between the two make sense?
The incompressible case is easier; I don't think it is stated in any source directly (because it is just known), but it is well known and I can tell you why. The incompressible case simplifies the physics and the space of degrees of freedom by assuming that $\nabla\cdot u = 0$. Physically, this eliminates lots of behaviors that are numerically difficult, most importantly shock waves. If you look at solvers for the compressible case (e.g. Riemann solver), a lot of work is spent managing these phenomena and accurately representing them, which is just not an issue here. Shocks are a particular problem because they introduce discontinuities and numerical conditioning issues; people have been struggling with this since the dawn of numerics, which is probably why finding a source that says "and we all know this case is harder" is hard.
Additionally, in many numerical schemes,$\nabla\cdot u = 0$ directly enables fast and efficient algorithms. The basics are that we represent the linearization of incompressible Navier-Stokes system as a block matrix where there is a velocity-velocity block $A$, velocity-pressure block $B$, pressure-velocity block $B^{T}$, and pressure-pressure block $C$ (but this is $C=0$ since pressure interacts through velocity):
$ X = \left[\begin{array}{ c | c } A & B \\ \hline B^{T} & C \end{array}\right] $
The consequence of $\nabla\cdot u = 0$ is that $B^{T}$ is easier to deal with. Since $C=0$, this means you have a system: $ X = \left[\begin{array}{ c | c } A & B \\ \hline B^{T} & 0 \end{array}\right] $
which almost looks like you could solve $B^{T}$ first and then solve the rest (Look up Schur complements). This is not too far from the truth as there are many schemes that in this case let you solve just with velocity and then recover pressure. This is only possible with incompressibility. More generally, we just have a good numerical idea how to deal with $B^{T}$ in the context of the this system.