When a cannonball is shot out of a cannon and the cannonball moves much faster than the cannon, does that violate Newton's third law?
It does not. Newton's third law says that the cannonball and cannon will exert equal and opposite forces on each other, but it does not directly address the acceleration, which is the key difference between the two objects' motions. Newton's second law holds that F = ma, and the force on each must be equal in magnitude (according to Newton's third law), but since the cannon has a much greater mass, it does not experience as large of an acceleration as the cannonball.
Why do you have to restrict the domain of inverse trig functions?
Since inverse trig functions spit out what you would put into the trig functions (various angles), the range of the inverse functions would consist of the domain of the trig functions. The catch is that when dealing with trig functions, you can put in many different angles to get the same resulting trig function value by completing another cycle around the unit circle (think sin(0),sin(2pi),sin(4pi), etc. are all equal to 0). Because of this, inverse trig functions would ordinarily spit out many different numbers if you wanted to find something like inverse sine of 0. This, however, violates what we think of as a function, which should have at most one y value for every x value. Thus, the ranges of the inverse trig functions get restricted to cut off all those extra values that would satisfy the relation, but create ambiguity when reversing the trig functions.
What's the difference between an indefinite integral and a definite integral?
The definite integral essentially applies some bounds to the more general indefinite integral, so that you can find the area under the curve between to x-values (with area under the x-axis counted as negative). You will employ one of the various methods of integration to find the indefinite integral of an expression, and if you want to find the area under the curve between two x-values, you find the difference between the values obtained when plugging in the bounds of integration into that expression. Remember, integrating an expression produces a constant of integration that only falls out when you specify the bounds of integration.