When integrating, is $C=0$ a unique solution?

Question

Why do we treat a certain $C$ as adding nothing?

take $\sin(x)$. It's usually agreed that: $$\int\sin(x)dx=-\cos(x)+C$$

But I could define a function like: $$\mathrm{sock}(x) = \cos(x)-1$$ And rightfully claim that $\int\sin(x)=-\mathrm{sock}(x)+C.$

And then $C=0$ is a different function, and in addition when integrating twice the result is very different, going to infinity instead of being bounded by $-1$ and $1$.

Basically, is there a reason that we choose $C=0$ to be the point that it is? This is also confusing from a physics standpoint, since even when choosing initial conditions yourself they will get you completely different effects/motions depending on the function that is considered the integral with $C=0$.

Answer 1

To understand this, go back to the basic definition of what an integral is. An indefinite integral (one with no limits) basically asks you to consider what function when differentiated will yield the function in the integral. That is the very heart of what an indefinite integral is.

But, we are then left with the following problem. If I was looking for the integral of cos(x), that could be any of the following

1. sin(x)
2. sin(x) + 1
3. sin(x) + 1000
4. sin(x) + 3.14159

The point here is any constant simply disapears when we integrate. Try it! all of the above function when differentiated will yield cos(x). So therefore, in order to represent all of the possible constants, we simply represent them with the letter C. The actual number that C represents in completely unimportant.

Answer 2

I wonder what you mean with that "$C=0$" (and especially "to be the point that it is???").

In indefinite integrals, the integration constant $C$ just remains unspecified. And indeed, if you integrate twice, you get a term $Cx+C'$. After $n$ integrations, an indeterminate polynomial of degree $n-1$. This is just another way to remind that a polynomial vanishes if differentiated $n$ times.

This causes no practical problem, because you don't mix indefinite and definite integrals so either the constant(s) remains indeterminate or cancel out. And context tells you which is appropriate.

In ODEs, the initial conditions do allow you to compute the relevant value of $C$, and this is equivalent to performing definite integration. In fact, no physical problem relies on indeterminate integrals.

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Anecdotically, the function $1-\cos x$ was used in the past an is known as the versine.

Answer 3

It's hard to know what you mean when you say "different function". Sure,

$$f(x)=-\cos(x)$$

and

$$g(x)=-\cos(x)+1 \equiv -\text{sock}(x)$$

are different functions in the sense that $f(x)\neq g(x)$ for all $x$. However, unless you're given an initial condition, you're not looking for one function when you integrate like this. You can't find one and only one function without more information. The $+C$ simply indicates that you're dealing with a "family" of functions: i.e. functions of the form $-\cos(x)+k$ where $k$ is some real number.

In terms of applications to physics, $C$ doesn't have to be zero at all. Of course, it can be helpful when $C$ is zero because you have no constant, but you don't need $C$ to be zero. It depends on the problem, but if you're dealing with something like projectile motion with an initial vertical displacement, your value of $C$ won't be zero. There's no general principle that says "we make $C=0$". It all depends on the problem and what you are given.

Answer 4

I'm not certain I fully understand the question as stated, but perhaps this example will shed some light on the role of the constant $C$ of integration, and in particular that it is meaningless without a fixed choice of antiderivative.

Consider the integral $$\int 2 \sin x \cos x \,dx .$$ One can solve this using the substitution $$u = \sin x, \qquad du = \cos x\,dx,$$ which yields $$\int 2 u \,du = u^2 + C = \sin^2 x + C .$$ Instead using the substitution $$v = \cos x, \qquad dv = -\sin x\,dx$$ gives $$-\int 2v \,dv = -v^2 + C = -\cos^2 x + D.$$

Both of the resulting expressions are antiderivatives of the original integrand, $2 \sin x \cos x$, but setting the constants $C, D$ of integration to zero yields two different functions. (Indeed, equating the two antiderivatives gives that $D = 1 + C$.) In particular, it is meaningless to "set $C = 0$" before you've picked a particular antiderivative $F(x)$.

Remark Instead rewriting the integral as $\int \sin 2x \,dx = \frac{1}{2} \cos 2 x + E = \frac{1}{2} (\cos^2 x - \sin^2 x) + E$ gives a third inequivalent expression when setting the constant $E$ to zero.

Answer 5

The symbol $$\int{f(x)}\,\mathrm{d}x$$ is an informal notation, meant to denote the following set: $$\{F:D\rightarrow\mathbb{R},\,F'=f\},$$ where $D$ is the domain of $f,$ and $F'$ denotes the derivative function of $F.$ In other words, you are trying to find functions $F$ that solve the equation $F'=f.$ So when we say $$\int\sin(x)\,\mathrm{d}x=C-\cos(x),$$ what we are informally saying is that $$\{F:\mathbb{R}\rightarrow\mathbb{R},\,F'=\sin\}=\{\cos+C:C\in\mathbb{R}\}.$$ There is no inconsistency in also saying that $$\int\sin(x)\,\mathrm{d}x=\operatorname{sock}(x)+C,$$ because $$\{\cos+C:C\in\mathbb{R}\}=\{\operatorname{sock}+C:C\in\mathbb{R}\}.$$