I have this Fourier trasformate $$\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx$$ and I have to prove this relation
$$\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx =2 \int_{0}^{1}(1-x^2)\cos(px) dx$$ Hint: prove that $\hat{f}(p)$ is real and even.
When I solve an exercise about Fourier antitrasformate, I usually analyse what happens for p>0 and p<0 and this is the proceeding that I'm following for this one.
If p>0, $\displaystyle \hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx$
If p<0, $\displaystyle \hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{ipx}dx$ (now with "p" I'm indicating the absolute value of p)
So I'd say that $\displaystyle\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-i|p|x}dx$
I also know that $\displaystyle\cos (px)=\frac{e^{ipx}-e^{-ipx}}{2}$
But now I don't know to to go on... Many thanks for your help!