Electronic – Integral of phasor

Phasors are expressed analytically as complex numbers, therefore in the phasor domain the usual current divider formula is:

\$ I_L = I_g \dfrac{R}{R+jX_L} \$

Hence, using the modulus operator on both members and using its properties you get:

\$ \left|I_L\right| = \left|I_g \dfrac{R}{R+jX_L}\right| = \left|I_g\right| \dfrac{\left|R\right|}{\left|R+jX_L\right|} = \left|I_g\right| \dfrac{\left|R\right|}{\sqrt{R^2+X_L^2}}= 2 \, mA \dfrac{10}{\sqrt{10^2+10^2}}= 2 \, mA \dfrac 1 {\sqrt 2} = \sqrt 2 \, mA \$

Superimposition is OK, but you made a mistake: \$I_0\$ is a DC component, so it cannot pass through a capacitor (as you say, we are analyzing steady-state). Therefore there is no point in computing the impedance, because it is apparent that the capacitor won't contribute to it (at DC it has infinite impedance, so in the parallel with R is completely negligible, if you want to use this POV).

Therefore \$I_0\$ will pass through R alone, that's why your book computes \$v_u'\$ simply as \$R I_0\$.

At the end you will have a DC component in \$v_u\$ due to \$I_0\$ and a sinusoidal component with frequency \$f\$ due to \$I_s\$, so your output voltage will have this form:

\$ v_u(t) = V_0 + A \cos(2 \pi f t + \phi) \$

EDIT

(in response to a comment)

The impedance of a capacitor in the phasor domain is

\$ Z = \dfrac{1}{j\omega C} = j \, \dfrac{-1}{\omega C} = j X_C \$

where \$X_C=\dfrac{-1}{\omega C}\$ is called the reactance of the capacitor.

If you want to make a comparison with what happens in DC circuits, i.e. with resistance, you should take the modulus of the impedance \$|Z|\$, which expresses intuitively how much the current flow is impeded when trying to flow in the capacitor.

A you can see: \$ |Z| = \dfrac {1}{\omega C} \rightarrow \infty \quad\textrm{as}\quad \omega \rightarrow 0 \$

And since \$I=\dfrac V Z \;\Rightarrow\; |I|= \dfrac{|V|}{|Z|} = 0 \$ when \$\omega=0\$