Does picking $C=0$ as a constant of integration result in a nominated anti-derivative?

Question

Introductory calculus students are often introduced to the "indefinite integral" or anti-derivative before actually doing integrals because it makes the FTC seem natural (by some rather sketchy notation). One learns symbols like $$\int f(x)dx,$$ which defines a class of functions indexed by $\mathbb R$ (complex numbers aren't needed for now), for instance $$\int x^2dx=\frac 1 3x^3+C\equiv\{\text{all such functions as }C\in\mathbb R\}.$$ However, it is common to forget the constant and think that the anti-derivative is just the primitive $F(x)=\frac 1 3x^3$. Given some sufficiently nice function $f$ and its class of anti-derivatives denoted $F$, is there some simple way of specifying the element of $F$ where the constant of integration is zero without referring to an arbitrary thing like $C$ (after all, you could have used $\tan C$ as $C\in(-\pi/2,\pi/2)$ or so on)?

Answer 1

The notion of "specifying the [particular antiderivative] where the constant of integration is zero" isn't well-defined: Usually there is no preferred choice of constant, and hence no preferred choice of antiderivative.

Consider the illustrative example

$$\int 2 \tan x \sec^2 x \,dx.$$

If we substitute $u = \tan x$, $du = \sec^2 x \,dx$, the integral becomes $$\int 2 u \,du = u^2 + C = \tan^2 x + C,$$ and setting $C = 0$ gives the particular antiderivative $F_1(x) := \tan^2 x$.

On the other hand, if we substitute $v = \sec x$, $dv = \sec x \tan x$, the integral becomes $$\int 2 v \,dv = v^2 + C = \sec^2 x + D.$$ and setting $D = 0$ gives the particular antiderivative $F_2(x) := \sec^2 x$, which is different from $F_1(x)$.

Of course, both $F_1$ and $F_2$ are perfectly good antiderivatives here, and as expected they agree up to a constant, in this case via the Pythagorean Identity $$\sec^2 x = \tan^2 x + 1.$$