Moment of inertia about a tilted axis I apologize if it seems foolish, I am a beginner.                      
Suppose I am given an object, whose moment of inertia along the x,z axes are known. Suppose it rotates around a tilted axis, say at angle $\theta$ with the x axis. Can I find out the moment of inertia by the following method?
I resolve the rotatory motion of the object into its x and z components- $\omega \cos \theta$ and $\omega \sin \theta$. I know its MOI along the x and z axis, so I find out the work done by the rotation along the x and z axis- $2W_x= I_x \omega^2 \cos^2 \theta$ and $2W_z= I_z \omega^2 \sin^2 \theta$. If the axis passes through the CM, I know the values of $I_x$ and $I_y$ readily. If not, I use the parallel axis theorem to find out $I_x$ and $I_z$. I think the total work done will be $W= W_x + W_y$, so from here I can find the total work done (I am not sure about this, but in analogy to translational motion, it makes sense. But the total work done is also $\frac12 I \omega^2$. So making them equal, I can find I (note- the $\omega^2$ term will cancel out. Will my method give the correct answer here, or do I just have to use the classic irritating method of integration?  
To be precise, I claim $$I^2= I_x \cos^2 \theta + I_z \sin^2 \theta$$
 A: So you have an object whose principal mass moment of inertia values $I_1$, $I_2$ and $I_3$ you know for the body-centered coordinates

Now you place this object on a tilted axis and rotate it about z

and you want to find out the mass moment of inertia $I_{zz}$ about z.
You correctly identified the rotational speed on the body coordinates as
$$ \begin{aligned} 
  \omega_2 &= \omega \sin \theta \\ 
  \omega_3 &= \omega \cos \theta 
\end{aligned}$$
and that these cause the following components of angular momentum
$$ \begin{aligned} 
  L_2 &= ( I_2 \sin \theta) \omega \\ 
  L_3 &= (I_3  \cos \theta ) \omega
\end{aligned}$$
except you went with energy instead of momentum, which might have obscured things because energy is a single scalar, and momentum is a vector and thus easier to understand what the terms represent (what orientation things are defined in).
The last step is to rotate the momentum components back into the world coordinate system
$$ \begin{aligned} 
  L_{yz} &= L_2 \cos \theta - L_3 \sin \theta  \\ 
  L_{zz} &= L_2 \sin \theta + L_3 \cos \theta 
\end{aligned}$$
which leads to the value of MMOI once the rotational speed is factored out
$$ \begin{aligned} 
  L_{y} &=  \underbrace{\left( (I_2-I_3)\sin \theta \cos\theta \right)}_{I_{yz} }  \omega \\ 
  L_{z} &= \underbrace{\left(  I_2 - (I_2-I_3)\cos^2\theta \right)}_{I_{zz} }  \omega
\end{aligned}$$
So your answer is
$$  \boxed{ I_{zz} = I_2 - (I_2-I_3)\cos^2\theta } $$

But it is a lot easier for me to do the transformation using the following matrix product, because it will produce all the components of the mass moment of inertia tensor.
Given the body inertia values as the following diagonal 3×3 matrix $$I_{\rm body} = \pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} $$
and the rotation about the x-axis by the angle $\theta$ as a 3×3 tranformation matrix $$ R = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta} $$
then the mass moment of inertia tensor in the world coordinates is given by the congruent transformation $I_{\rm world} = R I_{\rm body} R^\top$
$$ I_{\rm world} = \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}\pmatrix{I_1 & & \\ &I_2 & \\ & & I_3} \pmatrix{1 & 0 & 0 \\0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta}^\top $$
$$ \boxed{ I_{\rm world} = \pmatrix{ I_1 & 0 & 0 \\ 
0 & I_3 + (I_2-I_3) \cos^2 \theta & (I_2-I_3)\sin \theta \cos \theta \\
0 & (I_2-I_3) \sin \theta \cos \theta & I_2+(I_3-I_2)\cos^2 \theta } }$$
A: I have learnt that in 3D space if axis of rotation makes angle $\alpha$, $\beta$ and $\gamma$ from $x$, $y$ and $z$ axis respectively then moment of inertia can be calculated by $$I^2=(I_x\cos\alpha)^2+(I_y\cos\beta)^2 + (I_z\cos\gamma)^2$$
I hope that helps!
