I'm not quite sure what you're asking...but...first, in the world of complex variables trigonometric functions and exponential functions all unify. Thus by moving to complexes we have a nice uniform way of dealing with sine and cosine.
If $c_n=a_n+ib_n$, then
$$ c_ne^{int}+c_{-n}e^{-int} = c_ne^{int}+\bar{c_{n}}e^{-int} =$$ $$ (a_n+ib_n)(\cos(nt)+i\sin(nt))+(a_n-ib_n)(\cos(nt)-i\sin(nt)) = 2a_n\cos(nt)-2b_n\sin(nt) $$
So dealing with trig functions or exponentials is just a matter of notation/taste.
Now to address your question as to what these series are for...
Recall that an analytic function is equal to its Taylor series:
$$ f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k $$
If we stop this sum, say at $N$, we have
$$ f(x) \approx \sum\limits_{k=0}^N \frac{f^{(k)}(a)}{k!}(x-a)^k $$
which is a polynomial approximation of $f(x)$. Polynomials are nice. Taylor polynomials give us easily understood approximations of our function.
Now Fourier series do essentially the same thing for periodic functions. If we stop the summation at some $N$, we get a Fourier polynomial. This is a nice approximation made up of "easy" to understand trig functions.
From a theoretic viewpoint, Taylor series are wonderful because you can treat analytic functions sort of like polynomials. Fourier series allow one to treat nice periodic functions sort of like trig functions.
\begin{align}
a_0 &= \frac{1}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]}\, dx = \frac{1}{2\pi}\int_{0}^{\pi} \sin x \, dx
= \frac{1}{\pi}\\
a_n &= \frac{2}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]} \cdot \cos\tfrac{2\pi n x}{2\pi}\, dx
= \frac{1}{\pi} \int_{0}^{\pi} \sin x \cdot \cos nx \, dx
= \frac{1}{\pi} \frac{\cos (\pi n) + 1}{1-n^2}\\
&= \frac{1}{\pi}\frac{1 + (-1)^n}{1-n^2}
= \begin{cases} \frac{2}{\pi}\frac{1}{1-n^2} & \text{if $n$ even}\\ 0 & \text{if $n$ odd}\end{cases}\\
b_n &= \frac{2}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]} \cdot \sin\tfrac{2\pi n x}{2\pi}\, dx
= \frac{1}{\pi} \int_{0}^{\pi}\sin x \cdot \sin (nx)\,dx\\
&= \frac{1}{\pi}\cdot\begin{cases}\tfrac{\pi}{2} & \text{if $n=1$}\\0 & \text{if $n>1$}\\\end{cases}
= \begin{cases}\tfrac{1}{2} & \text{if $n=1$}\\0 & \text{if $n>1$}\\\end{cases}
\end{align}
Hence the Fourier series of $f(x)$ over $(0,2\pi)$ is given by
\begin{align}
f(x) &\sim
a_0+\sum_{n=1}^{\infty}\left [a_n \cos (nx) + b_n\sin(nx)\right]\\
&= \frac{1}{\pi} + \sum_{n=1}^{\infty}\frac{2}{\pi}\frac{1}{1-(2n)^2}\cos((2n)x) + \frac{1}{2}\sin((1)x)\\
&= \frac{1}{\pi} + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2nx)}{1-4n^2} + \frac{1}{2}\sin(x)
\end{align}
Plotted with 20 terms using Wolfram Alpha:
Note
I used Wolfram Alpha to compute these integrals, but it skipped the special case of $b_1$. This is likely to do with a symbolic division that implicitly assumed $n\neq 1$. Either way, this special case follows near-directly from the principle of orthogonality, i.e.,
$$
\frac{1}{\pi}\int_0^{2\pi} \sin(nx)\sin(mx)\,dx = \begin{cases}1 &\text{if $n=m$}\\
0 & \text{if $n\neq m$}\end{cases}
$$
So in our case, $$\frac{1}{\pi} \int_0^{2\pi} \sin(x)\mathbf{I}_{(0,\pi)}\sin(nx)\,dx
= \frac{1}{\pi} \begin{cases} \int_0^{\pi} \sin(x)^2\,dx & \text{for $n=1$}\\
\int_0^{\pi} \sin(x)\sin(nx)\,dx & \text{for $n>1$}\end{cases}$$
This latter point highlights why I knew that Wolfram Alpha hadn't given me everything,
$$\int_{0}^a \sin(x)^2 \,dx > 0,\quad\text{for $a>0$}$$
so we wouldn't expect the corresponding coefficient to be zero. If, on the other hand, the function were something like $f(x)=\sin(5x)\cdot \mathbf{I}_{(0,\pi)}$ or $f(x)=\cos(2x)\cdot \mathbf{I}_{(0,\pi)}$, then we'd know to check $b_5$ or $a_2$ respectively.
If you have learned about vector spaces, then you can think of the integral as an dot (inner) product over vectors $1, \sin(nx),\cos(nx)$ for $n=1,\dotsc,\infty$. The inner product measures similarity. In this case we are asking about amount of "$sin(x)$"-ness if we take a $\sin(x)$ function and lop off the latter half. The answer, perhaps unsurprisingly, is a half.
Best Answer
Use $c_{0} = a_{0}$ and solve the system
$c_{n}e^{inx}+c_{-n}e^{-inx} = a_{n}\cos\left(nx\right) + b_{n}\sin\left(nx\right)$
In the end you will get
$$ c_{n} = \frac{1}{2L}\int_{-L}^{L}f\left(x\right)e^{-inx}dx $$