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

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)?

• That would be wrong. In some cases you need that $C$. In the cases you don't need it e.g. defined integral it is because it is removed by the formula, but it is there nonetheless. – Lolman Mar 24 '15 at 11:38
• @Lolman Maybe I'm misunderstanding, but I'm aware that the constant is necessary (and cancels in the FTC). I'm asking whether the most natural primitive (with no constant) can be explicitly specified. – theage Mar 24 '15 at 11:40
• there is no natural primitive. that is the problem. There is a nice example with trigonometric functions. I think if you try by deriving 2 times the tangent function and integrate it one time you won't find an unequivocal way to write an answer. – Lolman Mar 24 '15 at 11:51

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 $$\sec^2 x = \tan^2 x + 1.$$

• Another example would be $\int 2 \sin(x) \cos(x) dx$ where you get the "natural" antiderivatives $\sin(x)^2$ and $-\cos(x)^2$. – Noiralef Mar 24 '15 at 11:58
• @Noiralef Yes, that's a very nice example, and perhaps better than mine as the involved functions are a little more familiar. Perhaps I avoided it in part because I answered a question about (very nearly) that example earlier today (math.stackexchange.com/questions/1204000/…) where the OP ran into some trouble with two particular antiderivatives that looked rather different. In fact, there is a third natural choice of antiderivative: using the double-angle identity $\sin 2x = 2 \sin x \cos x$ leads to the antiderivative $-\frac{1}{2} \cos 2 x$. – Travis Mar 24 '15 at 12:02