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I have to find Fourier coefficients of $x(t) = |A \cos (2 \pi f t )|$. The problem also give me $y(t) = |x(t)|$. To find Fourier coefficients I wrote
$$ \frac{1}{T_0} \int_{0}^{T_0} |A \cos (2 \pi f t )| e^{2 \pi k f t}$$
Solving this I found that $k$ should be equal to $1$ or $-1$. First I assume that $=1$ and I found the result $A/2 $ but on the book should be $2A/3\pi$. $$ T=\frac{T_0}{2} $$ Can someone help me?
Sketching the signal always helps in such cases. Let's consider a cosine of period $T_0$, that is, of fundamental frequency equal to $f_0=1/T_0$. That is the signal on the left.
As you can see on the right, $|A\cos(2\pi f_0 t)|$ does not share the same period with $A\cos(2\pi f_0 t)$. The period of the signal you are looking for is $T_0^{\prime}=T_0/2$ and its fundamental frequency is $f_0^{\prime} = 2f_0 = \frac{2}{T_0}$.
Assuming that you're looking for the exponential Fourier Series, the coefficients should be $$X_k = \frac{1}{T_0^{\prime}}\int_{T_0^{\prime}}x(t)e^{-j2\pi kf_0^{\prime}t}dt$$ and for $k=0$, which is usually calculated seperately, it is $$X_0 = \frac{1}{T_0^{\prime}}\int_{T_0^{\prime}}x(t)dt$$
Notice that you can integrate over any time interval of duration equal to $T_0^{\prime}=T_0/2$. I will pick $[-T_0/4, T_0/4]$ since the cosine is always positive in this interval and thus removing the $|\cdot|$ is trivial.
I will help with $X_0$ and you can do the $X_k$s. It is:
$$\begin{align} X_0 &= \frac{1}{T_0^{\prime}}\int_{T_0^{\prime}}x(t)dt = \frac{2}{T_0}\int_{-T_0/4}^{T_0/4}A\cos(2\pi f_0t) dt \\ &= \frac{2}{T_0}\frac{A}{2\pi f_0}\sin(2\pi f_0 t)\Big]_{-T_0/4}^{T_0/4} = \frac{2A}{2\pi f_0T_0}\Big(\sin\Big(\pi f_0\frac{T_0}{2}\Big) -\sin\Big(-\pi f_0\frac{T_0}{2}\Big)\Big) \end{align}$$
We know that $T_0 = 1/f_0 \Longrightarrow f_0T_0=1$ and that $\sin(-x)=-\sin(x)$. Simplifying:
$$\begin{align} X_0 &= \frac{A}{\pi}\Big(\sin\Big(\frac{\pi}{2}\Big) + \sin\Big(\frac{\pi}{2}\Big)\Big) = \frac{2A}{\pi} \end{align}$$
Now for the $X_k$s, I've already mentioned that you have to use $$X_k = \frac{1}{T_0^{\prime}}\int_{T_0^{\prime}}x(t)e^{-j2\pi kf_0^{\prime}t}dt$$ You can use the same interval for your integration, keep in mind that $f_0T_0=1$ to simplify your expressions, and treat $k$ as a constant until the end. Judging from your question, you should get something that includes $\frac{2A}{(2k+1)\pi}$.
If you are interested one more step into transform theory, I suggest you to go through Exponential Fourier series where the basis functions are complex exponentials, unlike sine and cosine as in Trigonometric Fourier series.
I will make an attempt to find the Fourier spectrum of $x(t) = A \cos (2\pi f_0 t)$ using Exponential FS and its properties(let me take $\omega = 2\pi f$):
$$\cos (\omega_0 t) = \dfrac{e^{j \omega_0 t}+e^{-j\omega_0 t}}{2}$$ Also it is quite easy to conceive the idea that FT of a signal that is spread over time will be concentrated in frequency domain. For e.g., $\mathcal{F}[1] = 2\pi \delta (\omega)$
Using frequency shift property of Exponential Fourier, $$\mathcal{F}[e^{j\omega_0 t}\times 1] = 2\pi \delta (\omega - \omega_0)$$ and $$ \mathcal{F}[e^{-j\omega_0 t}\times 1] = 2\pi \delta (\omega + \omega_0)$$
Therefore $$\mathcal{F}[\cos (\omega_0 t)] = \pi \delta (\omega - \omega_0) + \pi \delta (\omega + \omega_0)$$
The Fourier spectrum of this signal denoted by $X(j\omega)$ is shown in the figure below:
This figure clearly indicates that a Cosine signal possesses only one frequency component at $\omega = \omega_0$.
