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Homework Statement:
 Assume the period is 1/60 s. Given V(0) = `V(1/120) = V(1/60) = 0, and V(1/240) = V(1/80) = 100. Then output of a simple DC Generator will have the shape of absolute value of a sine function's curve. Expand this function in an appropriate Fourier series.
Relevant Equations:

Fourier series:
For a function with period ##2l##, can expanded in Fourier series:
$$f(x) = \frac{1}{2} a_0 + a_n \cos{\frac{n\pi x}{l}} + b_n \sin{\frac{n\pi x}{l}} $$
where a_n and b_n is Fourier's coefficient.
For even function (since absolute value is an even function), then $$b_n = 0$$ and $$a_n = \frac{2}{l} \int_0^{l} f(x) \cos{\frac{n\pi x}{l}} dx$$
First, I try to define the function in the figure above: ##V(t)=100\left[sin(120\{pi}\right]##.
Then, I use the fact that absolute value function is an even function, so only Fourier series only contain cosine terms. In other words, ##b_n = 0##
Next, I want to determine Fourier coefficient ##a_n##:
$$a_n=\frac{2}{l} \int_{0}^{l} V(t) \cos{120n\pi t} dt$$.
I think, now V(t) only positive, so ##V(t)=100 sin(120\pi t)##
then,
$$a_n=\frac{2}{l} \int_{0}^{l} \left(100 sin(120\pi t)\right) \cos{120n\pi t} dt$$.
At this stage, is my work true?
Can I evaluate the integral with Euler form? Thanks