Help in evaluating $\int_0^{\infty} \frac{2 \sin x \cos^2 x}{x e^{x \sqrt{3}}} dx$ I need some help in evaluating $ \displaystyle \int_0^{\infty} dx \frac{2 \sin x \cos^2 x}{x e^{x \sqrt{3}}}$
The original question: Evaluate $\displaystyle \int_0^{\infty} dx \frac{e^{- x \sqrt 3}}{x} (1 - \sin x)(1 + 2 \sin x - \cos 2x)$
Using $\cos 2x = 1 - 2\sin^2 x$ and $1 - \sin^2 x = \cos^2 x$ I was able to get it into the above form. However, I do not know how to proceed. I would  like some guidance rather than a full answer, please.
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\begin{align}
&\color{#c00000}{%
\int_{0}^{\infty}{2\sin\pars{x}\cos^{2}\pars{x} \over x \expo{x\root{3}}}\,\dd x}
=\int_{0}^{\infty}\expo{-x\root{3}}\,
{\sin\pars{x} + \sin\pars{x}\cos\pars{2x} \over x }\,\dd x
\\[3mm]&=\int_{0}^{\infty}\expo{-x\root{3}}\,
{\sin\pars{x} + \bracks{\sin\pars{x + 2x} + \sin\pars{x - 2x}}/2 \over x }\,\dd x
\\[3mm]&=\half\int_{0}^{\infty}\expo{-x\root{3}}\,
{\sin\pars{3x} + \sin\pars{x} \over x }\,\dd x
\\[3mm]&=\color{#c00000}{\half\int_{0}^{\infty}
\pars{\expo{-\root{3}x/3} + \expo{-\root{3}x}}\,
{\sin\pars{x} \over x }\,\dd x}\tag{1}
\end{align}

With $\mu > 0$:
  \begin{align}
&\color{#00f}{\int_{0}^{\infty}\expo{-\mu x}\,{\sin\pars{x} \over x }\,\dd x}
=\int_{0}^{\infty}\expo{-\mu x}\,\half\int_{-1}^{1}\expo{\ic kx}\,\dd k\,\dd x
=\half\int_{-1}^{1}\dd k\int_{0}^{\infty}\expo{\pars{\ic k - \mu}x}\,\dd x
\\[3mm]&=\half\int_{-1}^{1}{-1 \over \ic k - \mu}\,\dd k
=\half\int_{-1}^{1}{\ic k + \mu \over k^{2} + \mu^{2}}\,\dd k
=\int_{0}^{1}{\mu \over k^{2} + \mu^{2}}\,\dd k
=\color{#00f}{\arctan\pars{1 \over \mu}}
\end{align}

By replacing this result in $\pars{1}$ we find:
\begin{align}&\color{#00f}{\large%
\int_{0}^{\infty}{2\sin\pars{x}\cos^{2}\pars{x} \over x \expo{x\root{3}}}\,\dd x}
=\half\bracks{\arctan\pars{\root{3}} + \arctan\pars{\root{3} \over 3}}
\\[3mm]&=\half\pars{{\pi \over 3} + {\pi \over 6}}
= \color{#00f}{\large{\pi \over 4}} \approx 0.7854
\end{align}
A: I would consider the Laplace transform
$$F(p)=\int_0^{\infty} dx \, \frac{\sin{x}}{x} \cos^2{x} \, e^{-p x}$$
Then
$$\begin{align}F'(p) &= -\int_0^{\infty} dx \, \sin{x} \cos^2{x} \, e^{-p x}\\ &= -\frac14 \int_0^{\infty} dx \,(\sin{3 x}+\sin{x}) e^{-p x}\\ &= -\frac14 \left (\frac{3}{p^2+9}+\frac1{p^2+1} \right )\end{align}$$
Thus
$$F(p) = -\frac14 \left (\arctan{\frac{p}{3}}+\arctan{p} \right ) +C$$
The integration constant is
$$\begin{align}C &= \int_0^{\infty} dx \frac{\sin{x}}{x} \cos^2{x}\\ &= \frac14 \int_0^{\infty} dx \frac{\sin{3 x}}{x} + \frac14 \int_0^{\infty} dx \frac{\sin{x}}{x} \\ &= \frac14 \left ( \frac{\pi}{2} + \frac{\pi}{2}\right )\\ &= \frac{\pi}{4}\end{align}$$
so that
$$F(p) = \frac{\pi}{4} - \frac14 \left (\arctan{\frac{p}{3}}+\arctan{p}  \right ) $$
and your answer is $2 F(\sqrt{3})$ (which accounts for the factor of two in the original integral):
$$\begin{align}2 F(\sqrt{3}) &= \frac{\pi}{2} - \frac12 \left (\arctan{\frac{\sqrt{3}}{3}}+\arctan{\sqrt{3}}  \right )\\ &= \frac{\pi}{2} - \frac12 \left (\frac{\pi}{6} + \frac{\pi}{3} \right ) \\ &= \frac{\pi}{4}\end{align}$$
