I don't know this is eligible answer or not but I think it's worth being shared. This solution I made when trying to evaluate
$$
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx
$$
on Brilliant.org. These are the methods I use to evaluate that integral and post it there as a solution.
Method 1:
Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by
$$
\begin{align}
F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\
&=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\
&=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\
&=\frac{2a}{\omega^2+a^2}.
\end{align}
$$
Next, the inverse Fourier transform of $F(\omega)$ is
$$
\begin{align}
f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\
e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\
\frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.\tag1
\end{align}
$$
Now, rewrite
$$
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\mathbb{Re}\left(e^{2ix}\right)}{x^2+2^2}\,dx.\tag2
$$
Comparing $(2)$ to $(1)$ yield $t=2$ and $a=2$. Thus,
$$
\begin{align}
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx &=\frac{1}{2}\frac{\pi e^{-2\cdot|2|}}{2}\\
&=\frac{\pi}{4e^4}\\
\end{align}
$$
Method 2:
Note that:
$$
\int_{y=0}^\infty e^{-(x^2+4)y}\,dy=\frac{1}{x^2+4},
$$
therefore
$$
\int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx=\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx
$$
Rewrite $\cos2x=\Re\left(e^{-2ix}\right)$, then
$$
\begin{align}
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos2x}{x^2+4}\,dx\\
&=\frac{1}{2}\int_{-\infty}^{\infty}\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx\\
&=\frac{1}{2}\int_{y=0}^\infty\int_{x=-\infty}^\infty e^{-(yx^2+2ix+4y)}\,dx\,dy\\
&= \int_{y=0}^\infty e^{-4y} \int_{x=-\infty}^\infty \frac{1}{2} e^{-(yx^2+2ix)}\,dx\,dy.
\end{align}
$$
In general
$$
\begin{align}
\int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=-\infty}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
\end{align}
$$
Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then
$$
\begin{align}
\int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\
\end{align}
$$
The last form integral is Gaussian integral that equals to $\sqrt{\dfrac{\pi}{a}}$. Hence
$$
\int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).
$$
Thus
$$
\frac{1}{2}\int_{x=-\infty}^\infty e^{-(yx^2+2ix)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{(2i)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{1}{y}\right).
$$
Next
$$
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy.
$$
In general
$$
\begin{align}
\int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\
&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv.
$$
Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=v\;\rightarrow\;dt=dv$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.
$$
Thus
$$
\begin{align}
\int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
&=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\
\end{align}
$$
and
$$
\begin{align}
\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy\\
&=\frac{\sqrt{\pi}}{2}\cdot\sqrt{\frac{\pi}{4}}e^{-2\sqrt{4\cdot1}}\\
&=\frac{\pi}{4e^4}.
\end{align}
$$
It took me hours to make a solution using method 2 because almost no one on that site knows Fourier transform (it is understandable since most of the user there are only school students, so they know nothing about this method) so I made a solution using standard methods. Other methods to solve it is using contour integral or residual theorem, but I'm not familiar with those methods.
