How is the nuclear wavefunction norm defined in normal coordinates?

Lets assume we have a diatomic molecule with a total of six Cartesian coordinates. Lets also assume that the BO-Approximation is valid and we can write the ground state wavefunction like this,
$$
Psi_0(r,R) = chi(R)varphi_0(r,R)
$$
where $varphi_0(r,R)$ is the adiabatic electronic groundstate wavefunction. The nuclear wavefunction is $chi(R)$, still containing all nuclear degrees of freedom at this moment. The next step is separation of Rotation, Translation and Vibration. We have 1 vibrational degree of freedom since a diatomic molecule is linear. Lets assume that the system is in a stable minimum, so we can do a harmonic approximation for the electronic groundstate energy to solve for the vibrational eigenfunctions. This yields an harmonic oscillator hamiltonian for the vibrational wavefunction. We can write the wave function as
$$
Psi_0(r,R(Q)) = chi^{Rot,Trans}(Q_1,…Q_5)phi^{vib}_0(Q_6)varphi_0(r,R)
$$
where i have transformed to normal coordinates. In the harmonic approximation this transformation is linear. In this special case of a diatomic it is more or less superfluous since we have only one degree of vibrational freedom, but in general for systems with more nuclei we need this transformation to separate vibrational degrees of freedom from the rest.

The normalization condition should be
$$
int Psi^*_0(r,R)Psi_0(r,R)drdR = 1.
$$
The adiabatic electronic states are ortho normal at all $R$ by construction, so we can do the $r$ integration, which yields.
$$
int bar chi(R) chi(R) dR = 1
int bar chi^{Rot,Trans}bar phi^{vib}_0chi^{Rot,Trans}phi^{vib}_0det |frac{dR}{dQ}|dQ = 1.
$$

Now comes my question, can i split up this last integral into the following
$$
int det |frac{dR}{dQ}| bar chi^{Rot,Trans}chi^{Rot,Trans}dQ1…dQ5 int bar phi^{vib}_0 phi^{vib}_0(Q_6)dQ_6 = 1.
$$
and does it matter in any way where i put the differential volume factor ? The transformation to normal coordinates should be linear, at least approximately, so the factor should be a constant.