Question about FM and PM and Angle Modulation
The instantaneous frequency of a signal \$A\cos(\phi(t))\$ where \$\phi(t)\$ is an arbitrary function of time is defined as the derivative of \$\phi(t)\$ if you want to measure frequency in radians per second and as \$\frac{1}{2\pi}\$ times the derivaive_ of \$\phi(t)\$ if you want to measure frequency in Hertz.  Of course, in the common case of a
fixed frequency this corresponds to the familiar
\$\phi(t) = \omega_c t+\phi_0 = 2\pi f_c t + \phi_0\$.

The standard definition of a frequency-modulated signal is one in which the
deviation of the instantaneous frequency (at time \$t_0\$, say),
from the carrier frequency is proportional to the value \$x_m(t_0)\$ of the modulating signal \$x(t)\$ at time \$t_0\$.  The constant of proportionality is denoted by 
\$f_{\Delta}\$ in your notation: a \$1\$ volt signal creates a deviation of \$f_{\Delta}\$
Hz.  Thus, if \$A\cos(\phi(t))\$ is the FM signal, then we have that
$$ \left.\frac{\mathrm d}{\mathrm dt}\phi(t)\right|_{t=t_0} 
= 2\pi f_c + 2\pi f_{\Delta} x_m(t_0)
$$
so that the deviation of the instantaneous frequency \$f_c + f_{\Delta} x_m(t_0)\$
from the carrier frequency \$f_c\$ is \$f_{\Delta} x_m(t_0)\$, just as we want it to
be.  It then follows from the fundamental theorem of calculus that
$$\phi(t_0) = \int_{0}^{t_0}2\pi f_c + 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt
= 2\pi f_c t_0 + \int_{0}^{t_0} 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt$$
or, with a slight change in notation, the FM signal can be expressed as
$$A\cos\left(2\pi f_c t + \int_{0}^{t} 2\pi f_{\Delta} x_m(\tau)\, \mathrm d\tau\right)$$
the way you have it.  Note that \$A\$ is the amplitude of the FM signal and is
fixed; it is the frequency that is varying. Surely we need to distinguish
between the FM signal when it is created using a voltage-controlled
oscillator with an amplitude of \$1\$ volt and when it comes out of the
power amplifier and goes to the antenna with a power of 10 kW?