I accidentally bought a 7.4 volt servo when I should have bought a 5. Seeing that it is already integrated into the project it is being used for, what is the best way to drop 11 volts to 7 or 8 in a lightweight manner?
If by "lightweight," you mean "quick and dirty," then the best way is probably just to use a standard 3-terminal voltage regulator:
The LM317 adjusts the out terminal to 1.25V above the adjust terminal (which would be called ground in a traditional 7805 regulator). The purpose of the resistor network (now a set of fixed resistors in response to comments) is to generate a reference voltage 1.25 V below the desired output voltage (= 6.15V above ground): Approximately 5 mA of current will flow through R2 when when OUT is 1.25V above ADJ. The ADJ terminal sources less than 100 uA, so the voltage across R3 is about 6.4V. This is a couple hundred millivolts above the desired 7.4V, which is ok for this application. Originally, this schematic used a potentiometer, so that 6.15V could be easily generated without resorting to non-standard resistors or combinations of standard resistors.
The datasheet example circuit shows both input and output capacitors, however the text has this to say:
CI [C1] is required when the regulator is located an appreciable distance from power supply filter. CO [C2] is not needed for stability; however, it does improve transient response.
The output capacitor, C2, is there to supply charge between the time a large load is added to the output and when the reg's error amplifier responds. The servo is already designed to work over a fairly broad voltage range (because they normally run unregulated from a battery), so the transient response is not very important in this application.
If this circuit is meant to be powered by a battery, which has a very low output impedance, or the regulator is a short (wire) distance from the power supply, then C1 is also optional.