About power dissipation of capacitors and inductors & I-V phase difference

Question

For instance, The Art of Electronics (by Horowitz & Hill) states that

ideal capacitors cannot dissipate power, even though current can flow through them, because the voltage and current are 90° out of phase.

This reasoning has always baffled me since \$\small P = V \cdot I\$ and if we consider multiplying two sinewaves that are 90° out of phase, the resulting wave is not a flat zero line.

The fact that ideal capacitors and inductors store the power (later releasing it) seems to explain the phenomenon and make a lot of sense, but the phase explanation seems be the more common one.

As I see it the current "leading" the voltage across a capacitor is pretty easy to understand somewhat intuitively, but if the power dissipation would depend on it, both capacitors and inductors would instead dissipate half the power compared to resistors (assuming alternating voltage/current of course).

So, am I just getting something totally wrong or is the out-of-phase explanation flawed?

edit:

I finally found the error in my own reasoning. I somehow thought that multiplying two identical sinewaves with each other would result a wave that would be centered on the same axis as the original ones ie. also having positive and negative values. But with in-phase sinewaves the negative values of course line up and result in only positive values. Only when the waves (functions) are exactly 90 degrees out of phase is their product centered on zero axis and thus "spends" as much time on the negative side as on the positive.

Answer 1

If the voltage across a capacitor is described as sin(x), then the current through the capacitor (being 90 degrees out of phase) can be described as cos(x). The power is given by the product of voltage and current. For a capacitor, this is proportional to sin (x)*cos(x). This product is given by sin(2x)/2 which is a sine wave. Thus it is positive for half the time and negative for half the time. This means that power goes into the capacitor half the time (stored in the electric field) and that same power is released back to the circuit during the other half. The net power dissipated in the capacitor is zero.