Understanding the relation between instantaneous frequency and phase of a sinusoid
An analogy: a car is at $x=0$ at $t=0$ and moves with velocity $v$ (constant). The famous formula for this is:

\begin{equation}
    x = v \cdot t
\end{equation}

By integrating velocity $v$ one gets position $x$, by derivating position $x$ one gets velocity $v$.

Now, take position to be a point in a circle (for this is convenient to use radians instead of raw meters along the circumference, but this does not really matter). Take velocity to be how fast that point moves around the circle (for this is better to choose an unit like radians per second to make it independent of the circle radius, but this does not matter neither). This would be the phasor with constant rotational speed, a.k.a sine wave. So, we can reinterpret the formula like:

\begin{equation}
    \theta = \omega \cdot t
\end{equation}

And the same logic as above applies: By integrating frequency $\omega$ one gets phase $\theta$, by derivating phase $\theta$ one gets frequency $\omega$.

You can verify by dimensional analysis that they both are the same formula actually (well, actually, radian is said to be a dimensionless unit, something like a 'dozen' or a 'mole'... that is another story):

\begin{equation}
[Length] = [Length/Time] \cdot [Time]
\end{equation}

When you integrate with respect to time, one less $[Time]$ appear in the denominator. The derivative does the opposite.