Torques in Euler equation

The Euler equation is given by $$\mathbf I\dot{\boldsymbol \omega}+\boldsymbol\omega\times \mathbf I\boldsymbol\omega= \mathbf M.$$ Also see here. It explains that The expressions for the torque in the rotating and inertial frames are related by $$\mathbf M_{inert} = \mathbf Q \mathbf M$$ where $$\mathbf Q$$ is the rotation tensor (not rotation matrix), an orthogonal tensor related to the angular velocity vector by $${\displaystyle {\boldsymbol {\omega }}\times {\boldsymbol {v}}={\dot {\mathbf {Q} }}\mathbf {Q} ^{-1}{\boldsymbol {v}}}.$$ Isn't $$\mathbf M$$ just the torque we compute outside then transferred to one inside the inclined frame, i.e. using the Euler angle get the orientation of the body and express the torque with basis with the body frame. Why is it so complex in wikipedia, or I misunderstood?

The Euler equation

from the angular momentum $$L=I_{in}\,\omega_{in}$$

$$\Rightarrow$$

$$\frac{d}{dt} \left(I_{in}\,\omega\right)=M_{in}$$

with: $$I_{in}=R\,I_b\,R^T$$ $$\Rightarrow$$

$$L=I_{in}\,\omega_{in}=R\,I_b\,R^T\,\omega_{in}=R\,I_b\,\omega_b$$

R is the rotation matrix between inertial system (index in) and body coordinate system, (index b) ) we obtain:

$$\frac{d}{dt} \left(R\,I_{b}\,\omega_b\right)=M_{in}$$

$$\Rightarrow$$

$$\frac{d}{dt} \left(R\,I_{b}\,\omega_b\right)=R\,I_b\,\dot{\omega}_b+\dot{R}\,I_b\,\omega_b=M_{in}\tag 1$$

with $$\dot{R}=R\,\tilde{\omega}_b$$ in equation (1)

$$R\,I_b\,\dot{\omega}_b+R\,\tilde{\omega}_b\,I_b\,\omega_b=M_{in}\tag 2$$

multiply equation (2) from the left with $$R^T$$ we obtain (with $$R^T\,R=E$$ unity matrix):

$$I_b\,\dot{\omega}_b+\tilde{\omega}_b\,I_b\,\omega_b=R^T\,M_{in}\tag 3$$

with $$R^T\,M_{in}=M_{b}$$ and $$\tilde{\omega}\,u=\omega\times\,u$$ we get the EULER equations

$$\boxed{I_b\,\dot{\omega}_b+{\omega}_b\,\times \,(I_b\,\omega_b)=M_{b}}$$

or:

$$\boxed{I_{in}\,\dot{\omega}_{in}+{\omega}_{in}\,\times \,(I_{in}\,\omega_b)=M_{in}}$$

$$I_{in}=R\,I_b\,R^T$$