Introduction
There are 10 minutes left in the exam. You have done pretty well so far and get to the last problem. You hope it's doable in ten minutes. The question is something like this:
Find the F.T. of $ x(t) = sin(\pi t) $. Simply using the formula in the table will result in no points. You must prove the F.T. of x(t).
Without thinking, you rush into the problem using the definition of the Fourier Transform...
The Wrong Way
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt $
$ = \int_{-\infty}^{\infty}sin(\pi t)e^{-j\omega t} dt $
You don't know how to integrate that, but you remember that you can rewrite $ sin(\pi t) $.
$ = \int_{-\infty}^{\infty}\frac{e^{j\pi t}-e^{-j\pi t}}{2j}e^{-j\omega t} dt $
$ = \frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega - \pi)}dt-\frac{1}{2j}\int_{-\infty}^{\infty}e^{-jt(\omega + \pi)}dt $
$ = \frac{1}{-2jj(\omega -\pi)}e^{-jt(\omega - \pi)}|_{-\infty}^{\infty}-\frac{1}{-2jj(\omega + \pi)}e^{-jt(\omega + \pi)}|_{-\infty}^{\infty} $
$ = \frac{1}{2(\omega -\pi)}e^{-jt(\omega - \pi)}|_{-\infty}^{\infty}-\frac{1}{2(\omega + \pi)}e^{-jt(\omega + \pi)}|_{-\infty}^{\infty} $
And now you realize that you're in trouble because you have infinite terms as a consequence of the limtis of integration, and you can't solve the problem. The ten minutes is up and you receive 0 points for all your work, maybe 1 if you're lucky. Don't do this! You have to look ahead to see what to do.
