Evaluation of Definite Integral 
Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx$

$\bf{My\;Try::}$ Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{2\sin 3x\cos x\sin 2x}{x}dx$
So we get $$I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{(\sin4x+\sin 2x)\sin 2x}{x}dx$$
So $$I=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{\cos2x-\cos6x+1-\cos 4x}{x}dx$$
Now How can i solve after that , Help me
Thanks
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\begin{align}
\color{#f00}{I} & =
{1 \over 4}\int_{0}^{\pi/2}
{\cos\pars{2x} - \cos\pars{6x} + 1 -\cos\pars{4x} \over x}\,\dd x
\\[3mm] & =
{1 \over 4}\int_{0}^{\pi/2}{\cos\pars{2x} - 1 \over x}\,\dd x -
{1 \over 4}\int_{0}^{\pi/2}{\cos\pars{6x} - 1 \over x}\,\dd x -
{1 \over 4}\int_{0}^{\pi/2}{\cos\pars{4x} - 1 \over x}\,\dd x
\\[3mm] & =
{1 \over 4}\int_{0}^{\pi}{\cos\pars{x} - 1 \over x}\,\dd x -
{1 \over 4}\int_{0}^{3\pi}{\cos\pars{x} - 1 \over x}\,\dd x -
{1 \over 4}\int_{0}^{2\pi}{\cos\pars{x} - 1 \over x}\,\dd x
\\[3mm] & =
{1 \over 4}\braces{\vphantom{\Large A}%
\bracks{\vphantom{\large A}\mathrm{Ci}\pars{\pi} - \ln\pars{\pi} - \gamma} -
\bracks{\vphantom{\large A}\mathrm{Ci}\pars{3\pi} - \ln\pars{3\pi} - \gamma} -
\bracks{\vphantom{\large A}\mathrm{Ci}\pars{2\pi} - \ln\pars{2\pi} - \gamma}}
\\[3mm] & =
\color{#f00}{{1 \over 4}\bracks{%
\mathrm{Ci}\pars{\pi} - \mathrm{Ci}\pars{3\pi} - \mathrm{Ci}\pars{2\pi}
+ \ln\pars{6\pi} + \gamma}} \approx 0.8998
\end{align}
$\mathrm{Ci}$ is the Cosine Integral function and $\gamma$ is the Euler-Mascheroni constant.
A: Hint 1: $$ \int f(x) = F(X) + C \Longrightarrow \int f(ax+b) = \frac{1}{a} \cdot F(ax+b) + C$$
Hint 2: $$ \int \left(f(x) + g(x)\right) = \int f(x) + \int g(x) $$
Hint 3: $$ \int \frac{\cos x}{x} = Ci(x) + C$$
First step:
(use hint 2)
$$
4I=\int_{0}^{\frac{\pi}{2}}\frac{\cos2x-\cos6x+1-\cos 4x}{x}dx =\\
=2\int_{0}^{\frac{\pi}{2}} \frac{1}{x} + 2\int_{0}^{\frac{\pi}{2}}\frac{\cos 2x -1}{2x} - 6 \int_{0}^{\frac{\pi}{2}} \frac{\cos 6x -1}{6x}
- 4\int_{0}^{\frac{\pi}{2}} \frac{\cos 4x -1}{4x}
$$
Now let $f(x) := \frac{\cos x - 1}{x}$  so
$$
4I = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{x} + 2 \int_{0}^{\frac{\pi}{2}} f(2x) - 6 \int_{0}^{\frac{\pi}{2}} f(6x) - 4 \int_{0}^{\frac{\pi}{2}} f(4x)
$$
Edit after comment: You can also use Taylor expansion of $\cos x$. But then you will have series as result (without $Ci$ function).
