Sine and cosine waves are, physically, the most common. They are definitely the best description to what comes out of a wall socket, not because we like them mathematically, but because it's what comes out; electromotive force is generated in the power plant as a sinusoidal pattern with frequency 50/60 Hz. In the usual kind of generator, this is because in the generator, the rotating motion of the magnetic rotor leads to sinusoidal variation of EMF in the winding of the stator and consequently in any circuit connected to the socket. (Note that sine and cosine waves are equivalent, and choice between them is merely convention; the neutral word that can be used to describe the shape is harmonic.)
Even better, if we have some more complicated waveform - from slightly deformed all the way to ramp and step waveforms - then we can use Fourier series to decompose them into a sum of sine waves. We can then study the response of the circuit to each sine wave independently, and add up the responses at the end; this is usually much simpler than using the waveform directly.
The reason this works is because most circuits are linear. That is, if we input some voltage $v(t)$ and we measure some property $p(t)$ of the circuit, then adding a voltage $v'(t)$ will result in the property being $p(t)+p'(t)$. Thus if our complicated waveform $v(t)$ can be expressed as a Fourier series, say, as
$$v(t)=\sum_n v_n \sin(n\omega t),$$
and we know (because it's easier to study) that a sinusoidal stimulus $v_n\sin(n\omega t)$ will result in a property $p_n(t)$, then the full waveform $v(t)$ will induce a response
$$p(t)=\sum_n p_n(t).$$
This can, of course, break down if you have nonlinear elements in your circuit, such as diodes or overdriven vacuum tubes or transformers. These will behave differently and will induce distortion of your waveform, which may or may not be a good thing depending on what the application is. This distortion is, of course, the same distortion as you get with an electric guitar amp.
A final word on square and ramp waveforms. Fourier treatments of these waveforms are sometimes a bit difficult, and convergence near the sharp edges can be very slow (see e.g. the Gibbs phenomenon). This represents some very important physics of square waves: the instantaneous change in voltage is not actually possible in any physical circuit. This is because if the source circuit has any inductance $L$, the instantaneous change in current $i$ will mean an infinite current change $\frac{di}{dt}$ and therefore an infinite inductance back-voltage. Typical sources have very small inductances, but they can never be zero. The comparison of this inductance and the source internal resistance will give the timescale in which the voltage can suddenly change sign.
If you want a good description of an actual physical square wave, then, you have two choices: you can account for a finite "ramp" time, or equivalently you can simply take a finite number of terms in the Fourier series. The latter is, of course, a lot easier!
You are right, the transmission of AC does in fact produce EM waves around it. These waves are indeed sinusoidal, but we do not bother about the power losses until we start sending radio signals via these wires because, until the frequency is much higher than radio waves, the energy lost is extremely small and hence negligible. See
Ordinary electrical cables suffice to carry low frequency AC, such as mains power, which reverses direction 50 to 60 times per second. However, they cannot be used to carry currents in the radio frequency range or higher, which reverse direction millions to billions of times per second, because the energy tends to radiate off the cable as radio waves, causing power losses.
(SOURCE : http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Transmission_line.html)
Even for radio signals we do not lose a lot of energy but it start making a significant effect thereafter, this is why we do not consider the losses encountered as EM wave transmission while handling single/low frequency AC circuits.
When a square waves is passed into such a circuit it becomes a triangular wave , but why does the integrator circuit give the same output voltage graph when I pass a sinusoidal wave. Assuming time period is much smaller compared to the time constant of the circuit. If I pass a sine wave must not I get a cosine wave , why do I still get a sine wave
Best Answer
As said in the comments this may fit better in EE SE, nevertheless here is my answer:
Intuitively,a simple RC integrator circuit (without an Op-amp) works as long as you make sure that the voltage across the capacitor is small in amplitude relative to the input amplitude. In other words - as long as the capacitor can't keep up with the changes of the voltage the circuit works well. In order to make the capacitor unable to keep up we need to make sure that the RC constant of the circuit is large compared with the period of the input signal. If this condition is not maintained the circuit won't function as requested. It may be that the input signal violates this requirement, then the capacitor just follows the input signal (because when RC is small compared with the period the capacitor follows the input with little to no phase shift).
Also, how are you sure that the output of your circuit is a sine and not a cosine? make sure you compare the output of you circuit relative to the input, if you just measure them separately and watch them on your oscilloscope screen than the waveform will look the same because of how the triggering of the oscilloscope sweep works. You need to look at both input and output at the same time and than look for the phase shift.