Suppose, I have a function $\cos(x)$. Now,
$$\int{\cos(x)dx}$$
$$\sin(x)+c\\ [\text{Indefinite integral of cos(x)}]\\ [\text{where c is a constant}]$$
Now, there could be an infinite number of values for $c$. For example, $c=1,2,-2,\pi,-\pi, 0, \frac{1}{3}, 7, 500, ...$
And using each value of $c$, we can formulate an infinite number of definite integrals of $\cos(x)$:
$$\sin(x)+1[\text{Definite integral of cos(x)}]$$
$$\sin(x)+\pi[\text{Definite integral of cos(x)}]$$
$$\sin(x)+\frac{1}{3}[\text{Definite integral of cos(x)}]$$
$$...$$
So, $\cos(x)$ has an infinite number of definite integrals, hasn't it?
When the indefinite integral $\int{\cos(x)\,\mathrm dx}$ is instantiated as $\sin(x)+3,$ it becomes a particular antiderivative. An indefinite integral is represents every antiderivative of its integrand, and the $C$ in its specification is more accurately called an arbitrary constant or a parameter.
(If a function is defined on an interval and has an antiderivative, then its antiderivative is unique up to an additive constant and independent arbitrary constants aren't required in the general specification of its antiderivatives.)
Instantiating $\int{\cos(x)\,\mathrm dx}$ doesn't make it a definite integral, since the latter is defined with reference to specified integration limits (the $a$ and $b$ in $\int_a^bf\,)$ and denotes an area related to them.
(It's worth noting that a definite integral isn't actually a particular instance of an indefinite integral; for example, for \begin{align}f(x)&= \begin{cases} 0, &x\in[-1,0)\cup(0,1];\\ 1, &x=0,\end{cases}\end{align}$\int_{-1}^1 f(x)\,\mathrm dx=0$ even while $f$ has no antiderivative and so no indefinite integral.)
This function of $x$ is literally a definite integral.
If $f$ is a continuous function, then by the Fundamental Thereom of Calculus Part 1, $∫^x_0f(t)\,\mathrm dt$ is also an antiderivative of $f$ and, being a single antiderivative, isn't an indefinite integral.
P.S. Read more here: What does the antiderivative represent?