RC integrator: why does it convert a triangular wave into a sine wave?
So, a square wave has harmonics at all odd multiples of the fundamental, with strength diminishing as the number of the harmonic:
$$x_s(t) \simeq \sum_{n = 0}^\infty \frac{\cos 2 \pi (2n + 1) t}{2n + 1}$$
A triangle wave is just the square wave, integrated, with the appropriate constant added to make things tidy:
$$x_\Delta(t) \simeq \sum_{n = 0}^\infty \frac{\sin 2 \pi (2n + 1) t}{(2n + 1)^2}$$
A parabolic wave is the same thing, again:
$$x_p(t) \simeq \sum_{n = 0}^\infty -\frac{\cos 2 \pi (2n + 1) t}{(2n + 1)^3}$$
Because of that cubic in the denominator, the difference between a sine wave and this &quot;pseudo-sine&quot; wave is very small.  And it turns out that it's pretty hard to see the deviation on a graph anyway.
If you plot the quadratic pseudo-sine wave on a graph, superimposed on a real sine wave, you'll see the (slight) difference.