I want to represent the signal $e^{j \beta \cos 2 \pi f_m t}$ in terms of its Fourier Series and then represent the result in terms of Bessel's Function. I have computed $c_n$ and would like someone to check if what I have done is correct or not. Thanks for helping.
$\newcommand{\Real}{\operatorname{Re}}\newcommand{\dd}{\; \mathrm{d}}$ \begin{align*} c(t)&= A_c\cos (2\pi f_c t)\\ u(t)&= A_c\cos [2\pi f_c t + \beta \cos 2\pi f_c t]\\ u(t)&= \Real[A_c e^{j2\pi f_c t} e^{j \beta 2\pi f_c t}] \end{align*} Since $\cos 2\pi f_m t$ is periodic with period $T_m=\frac1{f_m}$, the same is true for complex exponential signal $e^{j \beta 2\pi f_m t}$. Therefore it can be represented as a Fourier series expansion.
Fourier series expansion is given by $$x(t)=\sum_{n=-\infty}^\infty c_n e^{j2\pi\frac{n}{T_0}t}$$ where $$c_n = \frac1{T_0}\int_{\alpha}^{\alpha+T_0} x(t) e^{-j2\pi\frac{n}{T_0}t} \dd t$$
\begin{align*} c_n &= \frac1{T_0}\int_0^{T_0} e^{j \beta \cos 2\pi f_c t} e^{-j2\pi\frac{n}{T_0}t} \dd t\\ &= \frac1{T_m}\int_0^{T_m} e^{j \beta \cos 2\pi f_c t} e^{-j2\pi\frac{n}{T_m}t} \dd t \\ &= f_m \int_0^{\frac1{f_m}} e^{j \beta \cos 2\pi f_c t} e^{-j2\pi f_mt} \dd t \tag{i} \end{align*}
Let $u=2\pi f_m t$, $\dd u=2\pi f_m \dd t$. $$ \begin{array}{c|c|c} t & 0 & 1/f_m \\ u & 0 & 2\pi \end{array} $$ $(i)$ becomes: \begin{align*} &= f_m \int_0^{2\pi} e^{j \beta \cos u} e^{-jun} \frac{\dd u }{2\pi f_m} &= \frac1{2\pi} \int_0^{2\pi} e^{j \beta \cos u} e^{-jnu} \dd u \tag{ii} \end{align*} Now $\sin[\pi/2+u]=\cos u$. $(ii)$ becomes $$=\frac1{2\pi} \int_0^{2\pi} e^{j \beta \sin [u+\pi/2]} e^{-jnu} \dd u \tag{iii}$$ $\sin [u+\pi/2] = \sin u \cos\frac\pi2+\cos u\sin\frac\pi2=\cos u$
Let $\pi/2+u=z$, $\dd u=\dd z$, $u=z-\frac\pi2$ $$ \begin{array}{c|c|c} u & 0 & 2\pi \\ z & \pi/2 & 5\pi/2 \end{array} $$ $(iii)$ Becomes \begin{align*} &=\frac1{2\pi} \int_{\frac\pi2}^{\frac{5\pi}2} e^{j \beta \sin z]} e^{-jn(z-\pi/2)} \dd z \\ &=\frac1{2\pi} \int_{\frac\pi2}^{\frac{5\pi}2} e^{j \beta \sin z]} e^{-jnz+jn\pi/2} \dd z \\ &=\frac1{2\pi} e^{jn\pi/2}\int_{\frac\pi2}^{\frac{5\pi}2} e^{j \beta \sin z]} e^{-jnz} \dd z \\ &=\frac1{2\pi} e^{jn\pi/2}\int_{2\pi} e^{j \beta \sin z]} e^{-jnz} \dd z \\ &=e^{jn\pi/2} J_n(\beta) \end{align*}
It's still a mess after Martin's edit, but you're essentially right. Your notation is inconsistent. I'll try to summarise.
We have $x(t) = e^{j\beta \cos (2\pi ft)}$. By a change of variables $\tilde t = 2\pi f t$, we can assume that $f = \frac1{2\pi}$ and $T=2\pi $, so we are tasked with computing $$ c_n :=\frac1{2\pi} \int_{0}^{2\pi} e^{j\beta \cos t} e^{-jnt} \, dt.$$ We seem to be happy with accepting the definition of the Bessel functions $J_n(\beta)$ as $$ J_n(\beta) := \frac1{2\pi}\int_{0}^{2\pi} e^{j\beta \sin t} e^{-jnt} \, dt.$$ Since $\cos t = \sin(t+\pi/2)$, changing variables $\tau = t+\pi /2$ gives $$ c_n = \frac1{2\pi} \int_{\pi/2}^{5\pi/2} e^{j\beta \sin \tau} e^{-jn(\tau - \pi/2)} \, d\tau = e^{jn\pi/2}\frac1{2\pi} \int_{0}^{2\pi} e^{j\beta \sin \tau} e^{-jn\tau} \, d\tau $$ where we used the fact that the integrand is periodic. By the way, $$e^{jn\pi/2}=i^n$$ this gives the formula in the Wikipedia page (Jacobi-Anger identity) linked in the above comment.