Derivation of Euler's equations for rigid body rotation 
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I do not understand from the line, "Now, in the body frame $T = (T_{x'}, T_{y'}, T_{z'})\ldots$"
How is that in the body frame? That should be in the inertial frame, since it is dL/dt.
In the body frame, it is dL'/dt. 
How can you treat dl/dt as the torque in the body frame, and then derive the Euler's equation?
 A: What is done here and maybe not stated so clearly is that the rate of change of angular momentum vector is broken up in two parts.

*

*Change in angular momentum due to angular acceleration $I\dot{\vec{\omega}}$

*Change in angular momentum due to inertia tensor rotation $\vec{\omega}\times I \vec{\omega}$
All this is explained in terms of a changing vector $\vec{L}$ riding on a rotating frame. Maybe this a good way to explain the equations, but not such a good way to derive them. To derive the equations follow this logic:

*

*There is a coordinate system attached to the body where the 3×3 inertial tensor is constant $I_{body}$. The 3×3 rotation matrix at some instant is $E(t)$.

*To find the 3×3 inertia tensor in world coordinates you need the transformation $$I=E(t)\,I_{body}\,E(t)^\top$$ with $^\top$ the matrix transpose. This is interpreted as a transformation to local coordinates, application of inertia tensor and transformation back to world coordinates.

*The body at the same instant has angular velocity $\vec{\omega}(t)$ described in world coordinates.

*The columns of the rotation matrix $E(t)$ are the unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$ of the coordinate system. Their time derivative equals only to a change in direction, as their magnitude is constant. So $$\frac{{\rm d}}{{\rm d}t} E(t) = \vec{\omega}(t) \times E(t)$$ where $\times$ is the vector cross product.

*The rate of change of the 3×3 inertia tensor is derived with the chain rule $$\begin{align}\frac{\rm d}{{\rm d}t} I(t) & = \frac{\rm d}{{\rm d}t} \left( E(t) I_{body} E(t)^\top\right) = \frac{{\rm d}E(t)}{{\rm d}t} I_{body} E(t)^\top + E(t) I_{body} \frac{{\rm d}E(t)}{{\rm d}t}^\top \\
 & = (\vec{\omega}\times E(t)) I_{body} E(t)^\top + E(t) I_{body} (\vec{\omega}\times E(t))^\top) \\
 & = \vec{\omega} \times I(t) - I(t) \vec{\omega} \times 
\end{align} $$ Don't worry about the 3×3 matrix cross product operator $[\vec{\omega}\times]$ as it is going be canceled out next.

*The angular momentum vector at the same instant is $\vec{L}(t) = I(t) \vec{\omega}(t)$ (by definition of the inertia tensor)

*Time derivative is angular momentum vector is derived with the chain rule $$\begin{align} \frac{{\rm d}}{{\rm d}t} \vec{L} &= I(t) \frac{{\rm d} \vec{\omega}(t)}{{\rm d}t} + \frac{{\rm d}I(t) }{{\rm d}t} \vec{\omega}(t) \\ &= I(t) \dot{\vec{\omega}} + \left( \vec{\omega} \times I(t) - I(t) \vec{\omega} \times \right) \vec{\omega} 
\end{align}$$
$$\boxed{\dot{\vec{L}}  = I(t) \dot{\vec{\omega}} + \vec{\omega} \times I(t) \vec{\omega} }$$
The last is Euler's equations of rotational motion.
The typo mentioned in comments is corrected.
NOTE: related answer https://physics.stackexchange.com/a/80449/392 for combined rotational and transnational equations of motion.
A: It clearly states that "...the body frame co-rotates with the body...". So what you are calling $L'$ is in fact $L$ itself. The torque will be the same in both reference frames.
