Let: $a>0$ $$ I(a) = \int^{\infty}_0 \frac{\sin(x)}{x}\cos(a\tan(x)) dx $$ Using Feynman's trick: \begin{align} I''(a)&=-\int^{\infty}_0 \frac{\sin(x)\tan^2(x)}{x}\cos(a\tan(x)) dx \\ &=-\int^{\infty}_0 \frac{\sin(x)(1+\tan^2(x))}{x}\cos(a\tan(x)) dx + \int^{\infty}_0 \frac{\sin(x)}{x}\cos(a\tan(x)) dx \end{align}
therefore: \begin{align} I''(a)-I(a)&=-\int^{\infty}_0 \frac{\sin(x)(1+\tan^2(x))}{x}\cos(a\tan(x)) dx \\ &=-\int^{\infty}_0 \frac{\sin(x)}{x\cos^2(x)}\cos(a\tan(x)) dx \end{align}
Let $f(x) = \cos(a \tan x)$. We have $f(\pi -x) = f(x)$. Thus, we can reduce the integral as
$$ \int_{0}^{\infty} \dfrac{\sin x}{x} \cos(a \tan x)\operatorname{d}\!x = \int_{0}^{\frac{\pi}{2}}\cos(a \tan x)\operatorname{d}\!x. $$ Let \begin{align} u &= \tan x \\ \operatorname{d}\!u &= \sec^2x\operatorname{d}\!x \\ \operatorname{d}\!x &= \dfrac{\operatorname{d}\!u}{1+u^2} \end{align}
The integral becomes $$ \int_{0}^{\infty} \dfrac{\cos(au)}{1+u^2}\,\operatorname{d}\!u =\Re\left(\int_{0}^{\infty}\dfrac{e^{iau}}{1+u^2}\,\operatorname{d}\!u\right). $$
Let us consider the complex function $f(z) = \dfrac{e^{iaz}}{1+z^2}$.
We choose a contour of the real axis from $-R$ to $R$ and a semicircular arc $(\Gamma_r)$ from $z = R$ to $z=-R$.
The poles are $\pm i$. But we will consider only $z=i$, as it is the only one inside our contour.
So, we have $$ \operatorname{Res}(f,i) = \lim_{z \to i}(z-i)f(z) = \lim_{z \to i} \dfrac{e^{iaz}}{z+i} = \dfrac{e^{-a}}{2i} $$
From here, we have $$ \int_{-\infty}^{\infty} \dfrac{e^{iaz}}{1+z^2}\,\operatorname{d}\!z = 2\pi i \dfrac{e^{-a}}{2i} = \pi e^{-a} $$
Since $\sin ax$ is odd, it's contribution is $0$, so we consider only $\cos ax$.
$$ \int_{-\infty}^{\infty}\dfrac{e^{iax}}{1+x^2}\,\operatorname{d}\!x = 2 \int_{0}^{\infty} \dfrac{\cos ax}{1+x^2}\,\operatorname{d}\!x = \pi e^{-a} $$
Thus, our result should be: $$ I(a) = \dfrac{\pi}{2} e^{-a} $$
I hope someone can point out my mistakes, if any.
Using real methods, $$I=\int\frac{\cos (a t)}{1+t^2} \,d t=\int\frac{\cos (a t)}{(t+i)(t-i)} \,d t$$ $$I=\frac i 2 \int \Bigg(\frac{\cos (a t)}{t+i } -\frac{\cos (a t)}{t-i } \Bigg)\,dt$$
Using an obvious change of variable and expanding the resulting cosine $$J(b)=\int \frac{\cos (a t)}{t+i b }\,dt$$ $$J(b)=\cosh (a b)\, \text{Ci}(a (i b+t))+i \sinh (a b) \, \text{Si}(a (i b+t))$$ Integrating from $0$ to $p$ and using asymptotics, assuming $a>0$, $$\color{blue}{\int_0^p\frac{\cos (a t)}{1+t^2} \,d t=\frac \pi 2 e^{-a}+\frac{\sin (a p)}{a p^2}-\frac{2 \cos (a p)}{a^2 p^3}-}$$ $$\color{blue}{\frac{\left(a^2+6\right) \sin (ap)}{a^3 p^4}+O\left(\frac{1}{p^5}\right)}$$ which is a good approximation.
If $a=1$ and $p=10$, converted to decimals, the above gives $0.574482$ while the exact value is $0.574239$ corresponding to a relative error of $0.04$%.
\begin{aligned} &\Omega=\int_0^{\infty} \frac{\sin x}{x} \cos (a \tan x) d x\\ &\text { Using Lobachevsky's identity since : } \cos (a \tan (\pi-x))=\cos (a \tan (\pi+x))=\cos (a \tan x)\\ &\Omega=\int_0^{\frac{\pi}{2}} \cos (a \tan x) d x=\int_0^{\infty} \frac{\cos (a t)}{1+t^2} d t=\frac{\pi}{2} e^{-a} \end{aligned}
