How does adding a cap between LM386 pins 1 and 5 boost bass frequencies?

I'm just starting out in Electrical Engineering and was wondering if anyone could clarify a question about the following RC circuit regarding the LM386.

My question is about the attached circuit:

How has adding such a small value cap (between pins 1 and 5) in parallel to the internal resistor of the op-amp boosted the bass?

I have been reading the data sheet; under gain control it says the following:

Additional external components can be placed in parallel with the internal feedback resistors to tailor the gain and frequency response for individual applications. For example, we can compensate poor speaker bass response by frequency shaping the feedback path. This is done with a series RC from pin 1 to 5 (paralleling the internal 15 kΩ resistor). For 6 dB effective bass boost: R . 15 kΩ, the lowest value for good stable operation is R = 10 kΩ if pin 8 is open.
But I do not understand how it works, as far as I knew the bigger the cap the less capacitive resistance it has, so wouldn't a small cap (0.033 µf) as in the diagram be more resistive to bass frequencies? I know I have misunderstood something here.

All help appreciated.

Your analysis is correct, it will and has to have a larger impedance at low frequencies. However, you need to think of how the whole feedback network behaves in analysing the thing.
Note that I am ignoring the fact that |Z|=sqr(r*r+Xc*Xc) throughout, does away with the need for a calculator. The assumptions also rely on the amplifier having an infinite gain bandwidth product.
Take the analysis to a limit, at f=0, Xc=infinity, or 15K//with an open circuit, which makes the feedback resistor appear to be 15K. At f=infinity, Xc=0, so we have 15K//10K, or 6K, and the feedback resistor is smaller, reducing the gain at high frequencies.
To get an idea of the value of the capacitor, assume that the resistor network has a value somewhere between 6K and 15K, and we want it to have more gain (base boost) at 100Hz.
If we assume a 10K impedance at 100Hz, this represents the internal 15K in parallel with say 30K,
c=1/(2*Pi*f*Xc), or 1/(2*3*100*30000), or 50nF, so we are in the right sort of area.
You can put the whole thing into spice if you want pretty graphs.
This was all done on the back of a piece of scrap paper, so all care but no responsibility, check it all out and see if you get similar results,
Cheers,
Richard

https://www.youtube.com/watch?v=sMZ7ew4LPlU&index=22&list=PLm3ZZSphEqeMojZkezqBpKNl-rerng6b7
Hope this may help.