Is work-energy theorem valid in non-inertial frames? From this answer for the question Is the energy conserved in a moving frame of reference?, I learnt that the work-energy theorem is independent of the frame of reference. But, is the theorem valid even for non inertial frames? I know that in non inertial frames we need to include inertial (pseudo or fictional) forces. Is the work done by all forces including the inertial forces equal to the change in kinetic energy?

Please note: According to this question and answer - Work associated with pseudo force, the work done by pseudo forces must be included in order to determine the total work done by all the forces. But the answer doesn't discuss about the validity of the theorem itself.
 A: Of course the theorem is still valid! To see why, let's review what goes into the proof of the work-energy theorem, which I'll state as
$$\Delta K = \int \mathbf{F} \cdot d \mathbf{x}, \quad K = \frac12 m v^2.$$
The easiest way to prove this is to work differentially,
$$dK = \mathbf{F} \cdot d \mathbf{x}.$$
By the definition of the differential,
$$dK = m \mathbf{v} \cdot d \mathbf{v} = m \frac{d\mathbf{x}}{dt} \cdot d \mathbf{v} = m \, d \mathbf{x} \cdot \frac{d \mathbf{v}}{dt} = m \mathbf{a} \cdot d \mathbf{x}.$$
Each of the steps here requires no physical input at all, besides the definitions of $K$ and $\mathbf{v}$, and some basic calculus like the product rule and the chain rule. So, we see the only physical assumption needed is
$$\mathbf{F} = m \mathbf{a}.$$
This is of course true in an inertial frame.
Now recall why fictitious forces are used. Going from an inertial reference frame to a noninertial one changes the acceleration. Therefore, naively it makes $\mathbf{F} = m \mathbf{a}$ stop working. The whole point of introducing fictitious forces is to adjust $\mathbf{F}$ so that $\mathbf{F} = m \mathbf{a}$ is true again. Then, as long as this holds, the proof of the work-energy theorem goes through exactly as above, so the theorem holds in non-inertial reference frames if you count the work done by the fictitious forces. 
A: In the following papers you can see the derivation and analysis of the work-energy theorem
in noninertial reference frames
Work and energy in inertial and noninertial reference frames
https://doi.org/10.1119/1.3036418
An extension to rotating reference frames is also given in
Work and energy in rotating systems
https://doi.org/10.1119/1.4807897
The work energy theorem is valid even for noninertial frames.
Best regards,
Diego Manjarrés
A: Yes why not, work energy theorem for a system of particles accelerating w.r.t an inertial frame is given as:-For an $n$ particle system let the inertial force on $i^{th}$ particle be$\vec{F_{inertial}}^i$ then,$$\sum_{i=0}^n \int (\vec{F_{inertial}}^i+\vec{F_{pseudo}}).d\vec{s}=\Delta K_{system}$$
$$W_{inertial}+W_{pseudo}=\Delta K_{system}$$ wheres change in mechanical energy of the system is given as $$\Delta E_{mechanical}=W_{ext}+W_{int,non cons}+W_{pseudo}$$
$W_{int,non cons}$ is the work done by internal non conservative forces of the system.
