With your new knowledge of capacitive reactance, you can see how when the input signal is an alternating current the capacitor and its reactance control the gain of the opamp.
With low frequencies, the reactance of the capacitor is high because a large current is stored that must be overcome each cycle in order for it to charge in the opposite polarity. Looking at the formula, you can see why this is true:
Xc = 1 / (2 pi f C)
With f approaching zero, the division gets larger and larger, approaching infinity.
One problem with having just a capacitor control the gain is that for low frequencies the gain can be so high as to drive the output to saturation on both polarities for each change in polarity of the input signal. To prevent this, a resistor is connected in parallel to the capacitor in order to limit the gain.
How this works is when the a low frequency is applied as input, the reactance of the capacitor will be extremely high, and since it is in parallel with the resistor, the equivalent resistance of the parallel combination will always be smaller than the smallest of values, so if the reactance is much higher than the resistance, so the resistance will dominate (when a component dominates is when a combination tends to the particular value of that component).
As the frequency at the input increases, the reactance of the capacitor decreases, making the parallel combination lower and lower. This has the effect that the ratio of Rin and Rf is smaller, making the gain of the amplifier lower and lower, given by the equation
Vout = Vin (-Rf / Rin)
With very high frequencies, Rf is dominated by the very low reactance of the capacitor, and the gain tends towards zero, so these frequencies are being blocked.
As you can see, the integrator circuit is also a low pass filter, amplifying low frequency signals and attenuating high frequency signals to the point of blocking them.
