Fourier series of a sinusoidal waveform: $E=E_{0}\cos(\omega t)$ without negative half-cycles. Statement of the problem: 
$a_{0}$ that I calculated is $\frac{E_{0}}{\pi}$. But I'm stuck with $a_{n}$ and $b_{n}$.
 A: First, in order to get some easier integrals, I suggest you try to solve this problem using the Exponential Form of the Fourier Series:
$$
\begin{align}
x(t)&=X_{0}+\sum_{n=-\infty}^{-1}\left[X_{n}\cdot e^{j\cdot n\cdot \omega \cdot t}\right]+\sum_{n=1}^{\infty}\left[X_{n}\cdot e^{j\cdot n\cdot \omega \cdot t}\right]\\
X_{0}&=\frac{1}{T}\int_{T}x(t)dt\\
X_{n}&=\frac{1}{T}\int_{T}x(t)\cdot e^{-j\cdot n\cdot \omega \cdot t}dt
\end{align}
$$
Indeed, as you stated, $X_{0}=\frac{E_{0}}{\pi}$
Now, for $X_{n}$:
Making $\omega=\frac{2\pi}{T}$, we can rewrite the function as $E_{0}\cos\left(\frac{2\pi}{T}t\right)$
We also know that, through Euler's, we have:
$$\cos\left(\frac{2\pi}{T}t\right)=\frac{e^{j\frac{2\pi}{T}t}+e^{-j\frac{2\pi}{T}t}}{2}$$
Then,
$$
\begin{align}
X_{n}=\frac{E_{0}}{T}\int_{0}^{T}\left(\frac{e^{j\frac{2\pi}{T}t}+e^{-j\frac{2\pi}{T}t}}{2}\right)\cdot e^{-j\cdot n\cdot \frac{2\pi}{T} \cdot t} dt
\end{align}
$$
Looking at the image you embedded, it is clear that the integral is not zero only for $t \in \left[\frac{T}{4};\frac{3T}{4}\right]$, so we say:
$$
\begin{align}
X_{n}&=\frac{E_{0}}{2T}\int_{\frac{T}{4}}^{\frac{3T}{4}}\exp\left(-j\frac{2\pi}{T}(n-1)t\right)+\exp\left(-j\frac{2\pi}{T}(n+1)t\right) dt\\
\end{align}
$$
Making:
$$
\begin{align}
a&=-j2\pi(n-1)\\
b&=-j2\pi(n+1)\\
\end{align}
$$
Then we have:
$$
\begin{align}
X_{n}&=\frac{E_{0}}{2}\left[\frac{\exp\left(\frac{3a}{4}\right)}{a}+\frac{\exp\left(\frac{3b}{4}\right)}{b}-\frac{\exp\left(\frac{a}{4}\right)}{a}-\frac{\exp\left(\frac{b}{4}\right)}{b}\right]\\
\end{align}
$$
Substituting $a$ and $b$, and doing all necessary algebra, keeping Euler in mind, you get something like this:
$$
\begin{align}
X_{n}&=\frac{E_{0}}{2\pi}\left[\frac{\cos\left(\frac{\pi n}{2}\right)}{\left(n-1\right)}-\frac{\cos\left(\frac{\pi n}{2}\right)}{\left(n+1\right)}\right]\\
&=\frac{E_{0}}{\pi}\left[\frac{\cos\left(\frac{\pi n}{2}\right)}{\left(n^2-1\right)}\right]\\
\end{align}
$$
Finally, your Fourier Series is this:
$$x(t)=\frac{E_{0}}{\pi}\left\{1+\sum_{n=-\infty}^{+\infty}\left[\frac{\cos\left(\frac{\pi n}{2}\right)}{\left(n^2-1\right)}\right]\cdot e^{j\cdot n\cdot \omega \cdot t}\right\}$$
For $n\neq0$
You can also notice that for any odd values of $n$, the argument of the sum is $0$.
As your original signal is real and even, it was already expected to get a real sum of cosines only!
Hope it was helpful.
A: WLOG, $E=1$ and $\omega=1$.
$$a_n+ib_n
\propto\int_{-\pi/2}^{\pi/2}\cos t\,e^{int}\,dt
\propto\int_{-\pi/2}^{\pi/2}\,e^{i(n+1)t}\,dt+\int_{-\pi/2}^{\pi/2}\,e^{i(n-1)t}\,dt=\left.\frac{e^{i(n+1)t}}{i(n+1)}\right|_{-\pi/2}^{\pi/2}+\left.\frac{e^{i(n-1)t}}{i(n-1)}\right|_{-\pi/2}^{\pi/2}\\
=\frac{e^{i(n+1)\pi/2}-e^{-i(n+1)\pi/2}}{i(n+1)}+\frac{e^{i(n-1)\pi/2}-e^{-i(n-1)\pi/2}}{i(n-1)}\propto\frac{\sin((n+1)\pi/2)}{n+1}+\frac{\sin((n-1)\pi/2)}{n-1}.$$
For odd $n$,
$$\frac{(-1)^{n+1}}{(n-1)/2}+\frac{(-1)^{(n+1)/2}}{n-1}=\frac{2(-1)^{(n-1)/2}}{n^2-1}$$
and for even $n$,
$$\frac{0}{n+1}+\frac{0}{n-1}=0.$$
Finally, the only nonzero coefficients are
$$a_{2m+1}\propto\frac{(-1)^m}{(2m+1)^2-1}.$$
