Consider the signal:
$\begin{align*} f(t) &= \sin(\omega t) \tag{$0 \leq t \leq 2\pi/\omega$}\\ &= 0 \tag{elsewhere} \end{align*}$
How to compute the Fourier transform of $f(t)$? I tried writing sin in terms of complex exponentials but it gets complicated. Thanks a lot for the help.
On
Writing it in terms of complex exponentials is definitely the way to go here. The Fourier transform of a function $f$ (when it exists) is given by
$$\mathcal{F}f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}f(t)\,dt.$$
I don't like that the frequency of your wave is $\omega$ so I'm going to change it to $\omega_0$ (this makes it seem like a fixed quantity rather than being variable). So what we want to evaluate is
$$\frac{1}{\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}e^{-i\omega t}\sin(\omega_0t)\,dt.$$
But since $\sin t = \frac{1}{2i}(e^{it}-e^{-it})$, we have that this becomes
$$\frac{1}{\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}e^{-i\omega t}\frac{1}{2i}(e^{i\omega_0t}-e^{-i\omega_0t})\,dt.$$
Rewriting this slightly gives
$$\frac{1}{2i\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}\left(e^{i(\omega_0-\omega)t}-e^{-i(\omega_0+\omega)t}\right)\,dt.$$
Here I am assuming $\omega\neq\pm\omega_0$. I will leave it to you to figure out these cases. We can do these integrals without too much difficulty:
$$\frac{1}{2i\sqrt{2\pi}}\left(\frac{1}{i(\omega_0-\omega)}e^{i(\omega_0-\omega)t}-\frac{1}{-i(\omega_0+\omega)}e^{-i(\omega_0+\omega)t}\right)\bigg|_0^{\frac{2\pi}{\omega_0}}.$$
Simplifying a bit gives
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{i(\omega_0-\omega)t}+\frac{1}{\omega_0+\omega}e^{-i(\omega_0+\omega)t}\right)\left.\right\lvert_0^{\frac{2\pi}{\omega_0}}.$$
Plugging in our limits gives...
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{i(\omega_0-\omega)\frac{2\pi}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{-i(\omega_0+\omega)\frac{2\pi}{\omega_0}}-\frac{1}{\omega_0-\omega}e^0-\frac{1}{\omega_0+\omega}e^0\right).$$
At this stage, you probably want to rearrange your exponents a little:
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{2\pi i-\frac{2\pi i\omega}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{-2\pi i-\frac{2\pi i\omega}{\omega_0}}-\frac{1}{\omega_0-\omega}-\frac{1}{\omega_0+\omega}\right).$$
Recognizing that $e^{2\pi i} = 1$, we get
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{-\frac{2\pi i\omega}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{\frac{-2\pi i\omega}{\omega_0}}-\frac{1}{\omega_0^2-\omega^2}\right).$$
Can you take it from here?
Fourier transform (FT) of one cycle of sine wave can also be obtained by using the FT of infinite cycle sine wave and the FT of a rectangular wave by using the multiplication property of the FT. Discussion below is just a technique. Kindly figure out how can you use the method below for your one cycle sine wave.
Consider the following functions and their FTs (From http://en.wikipedia.org/wiki/Fourier_transform)
$x(t)=\sin(\omega_0 t)$ -- FT$(x(t))=X(\omega)= \frac{\sqrt{2\pi}}{2 i} \left(\delta(\omega-\omega_0)-\delta(\omega+\omega_0 \right)$
$y(t)=rect(at)$--FT $[y(t)]=Y(\omega)=\frac{1}{\sqrt{2 \pi a^2}} sinc(\frac{\omega}{2 \pi a})$ -- $a$- width of the rectangular wave. We can take $a=2 \pi$ to cut one cycle of sine wave from the infinite cycle sine wave.
A one cycle sine wave $z(t)$ can then be created by the product $z(t)=x(t)y(t)$. Then, the FT of $z(t)$ is the convolution of $X(\omega)$ and $Y(\omega)$ (Multiplication property of FT).
Then $Z(\omega)=X(\omega)*Y(\omega)$ which can be given as
$Z(\omega)= \frac{1}{2 i\sqrt{a^2}} \left( sinc(\frac{\omega-\omega_0}{2 \pi a})-sinc(\frac{\omega+\omega_0}{2 \pi a}) \right)$
Please note that $X(\omega)$ is the FT of a single cycle sine wave (probably) whose domain may be from $-\pi/2$ to $+\pi/2$ (because the rectangualr function considered above it symmetric w.r.t zero). Please check it. If you want the result from 0 to $2 \pi$, additionally, please use the shifting theorem of FT to get your answer. I hope this helps you a bit.