Golden Rules. With negative feedback in place...
- No current flows into the input terminals of the opamp
- The voltage at the plus terminal equals the voltage at the minus terminal.
Because no current flows into the positive terminal, current must flow from the voltage source, through the series resistors 16 ohm and 24 ohm. So you should be able to figure out that the voltage at the input terminal is 7.5 * 24 / 40 = 4.5V. It follows that the voltage at the negative terminal = 4.5V. Now you've got another voltage divider from the negative terminal through 8 ohms in series with 12 ohms. The voltage across the 12 ohm resistor is therefore 4.5 * 12 / 20 = 2.7V.
Andy's comment is correct. It answers the question and explains what you are seeing, assuming you have swapped the + and - opamp inputs in your schematic : as drawn it has positive feedback instead of negative feedback!.
You are feeding the opamp with 12V, but there is no connection between that 12V supply and the 0V reference for your audio input, output, and scope probe.
Therefore you do not know whether it is +11V/-1V, +/-6V, or +1V/-11V.
And in fact it's probably all of these at different times. When it is +11V/-1V there isn't enough voltage on the -V supply rail for the opamp to function, and so the -ve peaks of the output are clipped.
If you had measured either of the supply rails as I suggested, you would see quite a large variation on them, probably a 50Hz noisy sinewave, whose peaks correspond to the distortion you see.
So you have to define the relationship between that 12V supply and your 0V. As a quick and crude test, take a pair of identical resistors, 1kilohm will probably do, and connect them between ground and +V, and between ground and -V. And observe with the scope probe that V+ is now about 6V (but probably still with a little noise on it).
As long as you aren't loading the opamp output with more than a couple of miliamps this will work, but in future, it's better to use +6V, 0V, and -6V supplies instead.
This arrangement with two resistors is often used on single supplies, but usually with a voltage regulator or even another opamp to provide a better "ground" than 2 resistors alone. It requires some care to remember that "0V" is not the negative supply rail!
Best Answer
No, you will get the same result in both cases, which is 0. You can see this is true by definition since the output is grounded.
Even with the output not grounded, flipping the capacitor won't make any difference. The polarity of a capacitor does not decide the polarity of the voltage applied to it. It only decides which polarity of applied voltage will cause the capacitor to blow up. This is assuming a type of cap to which polarity matters, such as electrolytic or tantalum.
In your circuit, the right side of the capacitor will always be at or lower than the left side due to the fact that only positive voltage is applied into the integrator. This particular integrator inverts as a side effect of its topology. Note that this circuit won't work unless the opamp has a negative supply. Without a negative supply, the opamp can't swing below ground and therefore can't make the expected output voltage.