Evaluating $\int_{-\infty}^\infty\frac{\cos(2x)}{x^2+4}\:\mathrm{d}x$ As stated in the title, I want to evaluate the integral
$$I=\int_{-\infty}^\infty\frac{\cos(2x)}{x^2+4}\:\mathrm{d}x$$
I'm pretty sure it evaluates to
$$\frac{\pi}{2e^4}$$
But I'm not sure how to evaluate it.
I have read an Instagram post where 3 different methods are provided for proving that
\begin{equation}
I(t)=\int_{-\infty}^\infty\frac{\cos(tx)}{x^2+1}\:\mathrm{d}x=\frac{\pi}{e^t}
\end{equation}
and I think similar logic can be applied here, but I am not sure how yet.
One of the methods mentioned in the post uses laplace transform to prove it but it's a little bit long. I'm wondering if there's any elegant method for evaluating $I$
I encountered this integral when I tried to solve this integral from one of the members of the Instagram math community.
$$\omega=\int_0^{\infty}\frac{x^2-4}{x^2+4}\:\frac{\sin 2x}{x}\mathrm{d}x$$
I first split the integral, used a property of laplace transform and some properties of the sine integral then used integration by parts and got to here
$$\omega=\frac{\pi}{2}-\left(2\int_{-\infty}^\infty\frac{\cos(2s)}{s^2+4}\:\mathrm{d}s+\pi\right)$$
Thank you so much for your help and attention! (BTW I'm not so proficient in complex analysis so I would prefer a solution without one :P)

I guess this might also help (no need to evaluate the derivative of Integral). The Fourier Cosine Integral is given by:
$$
f(w) = \int_{0}^{\infty }A(x)\cos(xw)dx
$$
where
$$
A(x) = \frac{2}{\pi}\int_{0}^{\infty }f(w)\cos(xw)dw
$$
Now let,
$$
f(w) = \frac{\pi}{2k}e^{-kw}
$$
Therefore,
$$
A(x) = \frac{2}{\pi}\int_{0}^{\infty }\frac{\pi}{2k}e^{-kw}\cos(wx)dw
$$
$$
A(x) = \frac{1}{k}\int_{0}^{\infty }e^{-kw}\cos(wx)dw
$$
In order to solve the above integral, we need the following equation (more on this can be found here):
$$
\int{e^{aw}\cos({bw})dw}= \dfrac{e^{aw}}{a^{2}+b^{2}}[a\cos(bw)+b\sin(bw)] + C
$$
If we substitute $a = -k$ and $b = x$, we get:
$$
\int_{0}^{\infty } e^{-kw} \cos(xw) dw = \frac{k}{x^2+k^2}
$$
Therefore,
$$
A(x) = \frac{1}{x^2+k^2}
$$
and therefore,
$$
f(w) = \int_{0}^{\infty }A(x)\cos(xw)dx
$$
can be rewritten as:
$$
\int_{0}^{\infty }\frac{\cos(xw)}{x^2+k^2}dx = \frac{\pi}{2k}e^{-kw}
$$
Now,
$$
\int_{-\infty}^{\infty }\frac{\cos(xw)}{x^2+k^2}dx = 2\int_{0}^{\infty }\frac{\cos(xw)}{x^2+k^2} = \frac{\pi}{k}e^{-kw}
$$
Now simply substitute w = 2 and k = 2 to get,
$$
\int_{-\infty}^\infty\frac{\cos(2x)}{x^2+4}\:\mathrm{d}x = \frac{\pi}{2e^4}
$$
