How does the Fourier Transform work??

Question

I was trying to understand how the Fourier Transform works and wanted to test it on $\cos(Ax)$. I know that the FT for $\cos(Ax)$ is $\delta(A)$ and $\delta(-A)$. So I wanted to check if for other values of $\omega$ if the FT is zeros. (Please note, I'm still ramping up on creating the math equation. I apologize if something is improper)

The FT of $\cos(x)$ (so $A = 1$) is: $$ F(\omega) = \int_{-\infty}^{\infty}\cos(x)e^{-jwt} dt $$ This implies, I'll have a $\delta(x)$ at $x=1$ and $-1$. For all other cases, it should be 0. $$ $$ Hence, at $\omega = 10$ (10 times the frequency of the original), the FT should be zero...

\begin{align} F(10) &= \int_{-\infty}^{\infty}{(e^{jt}+e^{-jt})e^{-10jt} dt}\\ &= {\frac{1}{2}(e^{-9jt})|_{-\infty}^{\infty}+ \frac{1}{2}(e^{-11jt})|_{-\infty}^{\infty}}\end{align}

But I'm stuck here and it looks like the answer is non-zero. From this page, I'm getting a non-zero result. Could someone please help in solving this integral? Also, please correct me if my assumption is totally wrong.