A more elegant way to find the Fourier transform Let  $f$ be defined analytically as : $$f(x)=\arccos \left ( \sin \left ( 2x \right ) \right ), x \in\left (0,10  \right ], f(x)=0, x\notin\left ( 0,10 \right ]$$
Here is a graph of the above function: 
The way I've come to solve this is to simply split the integral in the Fourier transform into pieces(the part where $f$ is non-zero), and do the calculations, but that is a tedious way of doing it. Do you have any suggestions how to retrieve the transform more quickly? 
EDIT: As I am only getting into Fourier analysis, I don't know many tricks. However I've been researching, and one interesting topic came up, the DFT. Is it applicable in this case?
Edit #2: Unfortunately, At the moment my time is very limited and I can't afford to go into too much detail. If someone has the answer in terms of the standard properties of this transform, or maybe even an idea different from what Jack suggested in his first comment, post it. I am really eager to see how this can be done. I suppose, we will learn about this in year 2 of college, in the Signals class.
EDIT #3: Is it possible to invoke some other transform and use it to obtain the Fourier transform?
 A: If (for whatever reason) you are able to see that the image under $\Delta=d^2/dx^2$ is ... what it is: a linear combination of $\delta$ and its derivative at $0$, but just multiples of $\delta$ at the other "corners", then you can use the distributional "symmetry" of $\Delta$ and the fact that $e^{i\xi x}$ is an eigenfunction:
$$
\int e^{-ix\xi}\cdot \hbox{your function $f$}\;dx
\;=\; \int {\Delta e^{-ix\xi} \over -\xi^2}\cdot f(x)\;dx
\;=\; {1\over -\xi^2} \langle e^{-ix\xi},\;\hbox{that image by $\Delta$}\rangle
$$
This is just a slightly dressed-up integration by parts, but (if one can see how things fit together) saves some work by cancelling several terms up-front.
A: Your graph consists of 
1) Triangles, which has FT of squares of (reshaped) sinc function;
2) Triangles then shifted infinitely many times, which corresponds to convolving with Shah function. Since the FT of Shah is another Shah, the FT of the the infinitely shifted triangles becomes a squared sinc multiplied by Shah, which is effectively a sampling.
3) Then it is multiplied by a square, which corresponds to convolving with a sinc in the frequency domain.
I am not sure if there is a nice closed form, but this seems to be a standard way of approaching it.
A: Alternatively, you could use the fourier transform of continuous triangular wave, multiply it by a rectangular wave in time doman, which would then become a sinc convolution in frequency domain.
