Are all constants of integration $C$ equal?

Question

This is a question from very (very) basic calculus, but it concerns indefinite integrals and the constant $ C $ we always add when we find the antiderivative. This concerns some problems with subtracting two indefinite integrals.

Say we have $ f(\theta) = \sin(\theta) $

And we have $ h(\theta) = \cos(\theta) $

(These are simple examples to illustrate the point.)

Now, $ \int \sin(\theta) d\theta = -\cos(\theta) + C $ And $ \int \cos(\theta) d\theta = \sin(\theta) + C $

This is the question:

Is $ (\int f(x) dx) - F(x) = (\int g(x) dx) - G(x) $?

Here, the capital letters represent the antiderivatives where $C = 0$.

Edit: This is a stupid question I posted in the very beginning before I even knew how to do things on MSE. Please take this question with a grain of salt. My real questions start on the (sadly) deleted account @AbysmalMathematics.

Answer 1

The answer to your question is definitely "No". Here is an example. I can write $\int 2 \sin x \cos x dx = \sin^2 x + C$. This means "any function whose derivative is $2 \sin x \cos x$ is equal to $\sin^2 x$ plus some constant". Also, however, $\int 2 \sin x \cos x dx = -\cos^2 x + C'$. If we treat $\int 2 \sin x \cos x dx$ as a well defined function, we can deduce $\sin^2 x + \cos^2 x = C' - C$; that is, $\sin^2 x + \cos^2 x$ is equal to an arbitrary constant, which it isn't. It's even worse if we "choose" $C=C'$, because $\sin^2 x + \cos^2 x \neq 0$. How we should interpret the pair of integrals is "any function whose derivative is $2 \sin x \cos x$ is equal to $\sin^2 x$ plus some constant and also equal to $-\cos^2 x$ plus some other constant", from which we can deduce that $\sin^2 x + \cos^2 x$ is constant, which is fine, because it is.

Answer 2

Conventionally, if F is the antiderivative of f, then $F(x)=\int _0 ^x f(t)dt=\int f(x)dx+C$ with $C=0$. Therefore, $\int f(x)dx-F(x)=C$ where C is the constant of integration of $\int f(t)dt$ that one can set as equal to any wanted (constant) value in $\mathbb{R}$.
In your situation, $\int f(t)dt -F(x)=\int g(t)dt -G(x)$ iff you choose the same constant of integration for $\int f(t)dt$ and for $\int g(t)dt$.
Hope this is clear enough !