Help to show if this is correct $\int_{0}^{\infty}{(e^{-2x}+2e^{-x})\cos^2(x/ 2)\sin(x)\over x}\mathrm dx={9\pi\over 16}$ We have this integral 

$$\int_{0}^{\infty}{(e^{-2x}+2e^{-x})\cos^2\left({x\over 2}\right)\sin(x)\over x}\mathrm dx={9\pi\over 16}\tag1$$

$\sin(2x)=2\sin(x)\cos(x)$ rearrange to get
$\cos^2\left({x\over 2}\right)={\sin^2(x)\over 4\sin^2\left({x\over 2}\right)}$ now substitute this into $(1)$
$$\int_{0}^{\infty}{e^{-2x}+2e^{-x}\over x}{\sin^3(x)\over 4\sin^2\left({x\over 2}\right)}\mathrm dx\tag2$$
$2u=x$
$$\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}{\sin^3(2u)\over 8\sin^2\left({u}\right)}\mathrm du\tag3$$
Using $\sin(2x)={2\tan(x)\over 1+\tan^2(x)}$ sub into $(3)$
$$\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}{8\tan^3(u)\over (1+\tan^2(u))^3}{1\over 8\sin^2(u)}\mathrm du\tag4$$
Simplified to
$$\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}{\tan(u)\cos^4(u)}\mathrm du\tag5$$
$$\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}{\sin(u)\cos^3(u)}\mathrm du\tag6$$
Using $\cos^3(u)={3\over 4}\cos(u)+{1\over 4}\cos(3u)$
$${3\over 8}\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}\sin(2u)\mathrm du+{1\over 4}\int_{0}^{\infty}{e^{-4u}+2e^{-2u}\over u}\sin(u)\cos(3u)\mathrm du\tag7$$
So far nothing is really  leading to proving this integral.
I need  help on how to show if this integral is correct or not.
 A: Let $$I(a)=\int_{0}^{\infty}{(e^{-2x}+2e^{-x})\cos^2\left({x\over 2}\right)\sin(ax)\over x}\mathrm dx.$$
Using
$$ \int_0^\infty e^{-ax}\cos(bx)\mathrm dx=\frac{a}{a^2+b^2}$$
one has
\begin{eqnarray}
I'(a)&=&\int_{0}^{\infty}(e^{-2x}+2e^{-x})\cos^2\left({x\over 2}\right)\cos(ax)\mathrm dx\\
&=&\frac12\int_{0}^{\infty}(e^{-2x}+2e^{-x})(1+\cos(x))\cos(ax)\mathrm dx\\
&=&\frac12\int_{0}^{\infty}(e^{-2x}+2e^{-x})\cos(ax)\mathrm dx+\frac12\int_{0}^{\infty}(e^{-2x}+2e^{-x})\cos(x)\cos(ax)\mathrm dx\\
&=&\frac12\bigg(\frac{2}{4+a^2}+\frac{1}{1+a^2}\bigg)+\frac14\int_{0}^{\infty}(e^{-2x}+2e^{-x})\bigg[\cos((1+a)x)+\cos((1-a)x)\bigg]\mathrm dx\\
&=&\frac12\bigg(\frac{2}{4+a^2}+\frac{1}{1+a^2}\bigg)+\frac12\bigg[\frac{1}{(a-1)^2+4}+\frac{1}{2 ((a+1)^2+1)}+\frac{1}{(a+1)^2+4}+\frac{1}{2 \left((a-1)^2+1\right)}\bigg],
\end{eqnarray}
and hence
\begin{eqnarray}
I(1)&=&\frac12\int_0^1\bigg(\frac{2}{4+a^2}+\frac{1}{1+a^2}\bigg)\mathrm da+\frac12\int_0^1\bigg[\frac{1}{(a-1)^2+4}+\frac{1}{2 ((a+1)^2+1)}+\frac{1}{(a+1)^2+4}+\frac{1}{2 \left((a-1)^2+1\right)}\bigg]\mathrm da\\
&=&\frac12(\frac{\pi}{4}+\arctan\frac12)+\frac12(\pi+4\arctan 2)\\
&=&\frac1{16}(3\pi+8\arctan\frac12+4\arctan 2).
\end{eqnarray}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}{%
\pars{\expo{-2x} + 2\expo{-x}}\cos^2\pars{x/2}\sin\pars{x} \over x}\,\dd x =
{9 \over 16}\,\pi:\
{\large ?}}$.

\begin{align}
&\int_{0}^{\infty}{%
\pars{\expo{-2x} + 2\expo{-x}}\cos^2\pars{x/2}\sin\pars{x} \over x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
\pars{\expo{-2x} + 2\expo{-x}}{1 + \cos\pars{x} \over 2}
{\sin\pars{x} \over x}\,\dd x
\\[5mm] = &
{1 \over 4}\int_{0}^{\infty}
\pars{\expo{-2x} + 2\expo{-x}}\,{2\sin\pars{x} + \sin\pars{2x} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{0}^{\infty}
\pars{\expo{-2x} + 2\expo{-x}}\,{\sin\pars{x} \over x}\,\dd x +
{1 \over 4}\int_{0}^{\infty}
\pars{\expo{-x} + 2\expo{-x/2}}\,{\sin\pars{x} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}\bracks{%
\int_{0}^{\infty}\expo{-2x}\,{\sin\pars{x} \over x}\,\dd x +
{5 \over 2}\int_{0}^{\infty}\expo{-x}\,{\sin\pars{x} \over x}\,\dd x +
\int_{0}^{\infty}\expo{-x/2}\,{\sin\pars{x} \over x}\,\dd x}\label{1}\tag{1}
\end{align}

Note that

\begin{align}
&\left.\int_{0}^{\infty}\expo{-\alpha x}\,{\sin\pars{x} \over x}\,\dd x
\right\vert_{\ \alpha\ >\ 0} =
\int_{0}^{\infty}\expo{-\alpha x}\,{1 \over 2}\int_{-1}^{1}\expo{\ic k x}
\,\dd k\,\dd x =
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}
\expo{-\pars{\alpha - \ic k}x}\,\dd x\,\dd k
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}{\dd k \over \alpha - \ic k} =
{1 \over 2}\int_{-1}^{1}{\alpha + \ic k \over k^{2} + \alpha^{2}}\,\dd k =
\int_{0}^{1/\alpha}{\dd k \over k^{2} + 1} =
\arctan\pars{1 \over \alpha}
\end{align}

\eqref{1} becomes

\begin{align}
&\int_{0}^{\infty}{%
\pars{\expo{-2x} + 2\expo{-x}}\cos^2\pars{x/2}\sin\pars{x} \over x}\,\dd x =
{1 \over 2}\bracks{%
\arctan\pars{1 \over 2} + {5 \over 2}\,\arctan\pars{1} + \arctan\pars{2}}
\\[5mm] = &\
{1 \over 2}\pars{{\pi \over 2} + {5 \over 2}\,{\pi \over 4}} =
\bbx{{9 \over 16}\,\pi}
\end{align}

Note that
  $\ds{\arctan\pars{x} =
{\pi \over 2}\,\mrm{sgn}\pars{x} - \arctan\pars{1 \over x} \implies
\arctan\pars{1 \over 2} + \arctan\pars{2} = {\pi \over 2}}$.

