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

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).
