AC through a pure inductor I've studied the AC circuit for an ideal inductor in many physics books. After deriving the final equation for current the integration constant $C$ is assumed to be $0$ by giving inadequate reasons. In this question I seek for an adequate reasons. 
Suppose an ideal AC voltage source is connected across a pure inductor as shown:  

The voltage source is$$V=V_0\sin(\omega t).$$  From Kirchhoff’s loop rule, a pure inductor obeys
 $$V_0 \sin(\omega t)=L\frac{di}{dt},$$ so
$$\frac{di}{dt}=\frac{V_0}{L} \sin(\omega t)$$
whose solution is
$$i=\frac{-V_0}{\omega L}\cos(\omega t)+C$$
Consider the (hypothetical) case where the voltage source has zero resistance and zero impedance.

In most of  elementary physics books $C$ is taken to be $0$ for the case of an ideal inductor. 

$$\text{Can we assume that } C \neq 0?$$

  
*
  
*(To me this is one of the inadequate reasons). This integration constant has dimensions of current and is independent of time. Since source has an emf which oscillates symmetrically about zero, the current it sustain also oscillates symmetrically about zero, so there is no time independent component of current that exists. Thus constant $C=0$.
  
*(Boundary condition) there might exist a finite DC current through a loop of wire  having $0$ resistance without any Electric field. Hence a DC current is assumed to flow through the closed circuit having ideal Voltage source and ideal inductor in series if the voltage source is acting like a short circuit like a AC generator which is not rotating to produce any voltage. When the generator starts, it causes the current through the circuit to oscillate around $C$ in accordance with the above written equations.  

 A: The fact is, in the context of ideal circuit theory, the inductor voltages are equal in the circuits below:

In the lower circuit, the inductor current has a constant component, i.e., the lower circuit is equivalent to your $C \ne 0$ case.
But, there's nothing remarkable or surprising about this.  Is there something else to your question that I'm missing? 


[I] am asking that why in in elementary physics text books the author
  directly writes C=0 for ideal inductor without mentioning the initial
  conditions?

The initial conditions are mentioned when the context is transient analysis.  For example, from Wikipedia:

However, when the context is AC (sinusoidal) circuit analysis, the underlying assumptions are (at least):
(1) All sources are sinusoidal and of the same frequency
(2) The circuit is in sinusoidal steady state, i.e., all transients have decayed.
When these conditions hold, can we use voltage and current phasors and the notion of impedance to analyze circuits.
A: The general idea behind this sort of derivation is that, with precious few exceptions, the near-perfect conductors that we have in the lab do have some finite resistance. It is usually fine to ignore such a small resistance in the steady-state regime, but it is crucial in killing off transients.
For a real system you will have an equation of the sort
 $$V_0 \sin(\omega t)=L\frac{di}{dt}+Ri,$$ 
whose full solution is
$$
i(t)=\frac{V_0}{R^2+L^2\omega^2}\left(R\sin(\omega t)-L\omega\cos(\omega t)\right) +C e^{-\frac{R}{L}t}.
$$
As you can see, taking the limit $R\to\infty$ after solving the equation is a slightly more complicated affair. The important aspect is that the limit affects both terms differently.
The steady-state term does not change much. In the well-defined limit that $R\ll L\omega$, the first term of this equation converges smoothly to the simpler solution $-\frac{V_0}{L\omega}\cos(\omega t)$. If there is a small, finite resistance, then the real solution will have a slightly different amplitude and phase, but if $R$ really is very small then the changes will be hard to detect. 
The transient term, on the other hand, is much trickier to take, as there isn't another resistance to compare $R$ to. In particular, there are two limits you can take, $R\to0$ and $t\to\infty$, and they produce different results depending on the order you take them in.


*

*If you leave $R$ fixed and take the limit of long times, then the exponential will die off if $R>0$, regardless of how small it is. This is the usual regime we work in, as the relaxation times $\tau=L/R$ of usual systems tend to be quite small. Thus, you can see the condition of setting $C$ to zero as saying that the solution also includes a transient current but that the relaxation time is much shorter than the times we'll actually be observing the system at, and our temporal resolution there.

*On the other hand, if you leave $t$ fixed and you take $R$ to zero, then you're in the regime where you are observing the transients. In that case, though, you probably do care about having the full waveform of the transient, and you want to leave the exponential untouched.

On the other hand, it is possible to conceive of circuits where the resistance throughout the whole circuit is in fact zero. You might do this, for example, by connecting your superconducting inductance to one terminal of a superconducting transformer, using superconducting wire.

For a circuit like this, in general, you will observe a DC current going around the superconducting loop with no appreciable decay, and your argument is correct. However, these are the lengths you have to go to, to get your argument working on the real world.
A: It's unclear what you are really asking, but it seems to be related to the constant C you got when doing the integral to find the current thru a inductor given it is driven with a sine voltage.
Stop and think what this equation is really trying to tell you.  If you consider current out as a function of voltage in, a inductor is a integrator.  Like any integrator, the initial condition needs to be specified.  That is all C is.  Put in this context, C is the initial current thru the inductor.  If a ideal inductor starts out with 1 A thru it when the sine voltage is applied to it, that average 1 A will remain with the cosine current added to it.
Real inductors (except very specialized superconducting ones) have a non-zero resistance.  This resistance can be thought of as being in series with the inductance for the purpose of circuit analisys.  That resistance will cause any existing current to exponentially decay to zero.  Therefore, after enough time, C is 0 in practical cases.  Also, usually the initial current is 0 anyway, so most of the time C even starts as 0.
