How do I find the Fourier transform of $f(x)$ given by:
$$f(x)=\begin{cases}1-x^2\quad -1\leq x\leq 1\\0\hspace{37.5pt} \lvert x\rvert>1\end{cases}$$
By using the definition of Fourier transform I get the integral
$$\frac{1}{\sqrt{2\pi}}\int_{-1}^{1} e^{-iwx} \,dx -\frac{1}{\sqrt{2\pi}}\int_{-1}^{1} x^2e^{-iwx} \,dx$$
but do not know how to solve this type of integral.
Those should be straight forward. The ant-derivative of $e^{i\omega x}$ is $\frac{1}{i\omega} e^{i\omega x}= -\frac{i}{\omega}e^{i\omega x}$. Evaluating between -1 and 1, $-\frac{i}{\omega}\left(e^{i\omega}- e^{-i\omega}\right)$. We can write $e^{i\omega}= cos(\omega)+ i sin(\omega)$ and $e^{-i\omega}= cos(-\omega)+ isin(-\omega)= cos(\omega)- isin(\omega)$ so $\frac{i}{\omega}\left(e^{i\omega}- e^{-i\omega}\right)= $$\frac{-i}{\omega}\left(cos(\omega)+ isin(\omega)- cos(\omega)- cos(\omega)+ i sin(\omega)\right)=$$ \frac{2i}{\omega}(-2i sin(\omega))= \frac{4}{\omega}sin(\omega)$
In fact it would have been simpler to use $e^{i\omega x}= cos(\omega x)+ i sin(\omega x)$ to begin with.
For $\int_{-1}^1 x^2 e^{i\omega x}dx$ use "integration by parts" Letting $u= x^2$ and $dv= e^{i\omega x}dx$. That will result in an integral $\int_{-1}^1 xe^{i\omega x}dx$ which you also do by "integration by parts", letting $u= x$, $dv= e^{i\omega x}dx$.