I'm the author of application for android called Calculator++ One of the users submitted a bug in the integration of trigonometric functions. And I'm absolutely disagree with him.
In my app I use approach to calculate trigonometric functions differently in different modes (I mean DEG/RAD modes), e.g.:
- in rad mode you can type $\sin(\pi/ 3)$ and get result $0.866$
- in deg mode you type $\sin(60°) (≡ sin(60))$ and get the same result
So, strictly speaking in different modes I've got different sin functions - in deg it is a function of degree argument and in rad mode - function of radian argument.
Calculator++ allows to get the integral of sin function and in both modes: $\int\sin(x)dx = - \cos(x)$
And here is the misunderstanding: user says that correct integration formula is:
- RAD: $\int\sin(x)dx = - \cos(x)$
- DEG: $\int\sin(x)dx = - \pi / 180 \times \cos(x)$ (user assumes that variable modification is needed for correct integration)
I want to ask who is correct in this argument - me or user?
I don't think there's right or wrong there; it depends on how you describe what the trigonometric functions and the integration are supposed to do. Oddly, I would a priori tend towards your viewpoint, but based on your statement "strictly speaking in different modes I've got different sin functions" and on your equation "sin(60°) (≡ sin(60))", I'd say the user has a point: If you really do consider these as different functions, then clearly these different functions have different derivatives and anti-derivatives. (Speaking of anti-derivatives, both of your options are wrong since they're missing the integration constant.)
However, it seems to me that the most natural view isn't that these are two different functions (one of which never appears anywhere in mathematics), but that the same function is being applied to the same argument, just expressed in different units. Thus, I wouldn't drop the $^\circ$ symbol, but would consistently use it to denote an angle unit of $\pi/180$. Then the DEG mode would correspond to always implicitly adding a degree symbol to the current number before applying a trigonometric function. Then you have just one function, and the integral of that function is of course the one you give in your first option (plus a constant).