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Vì ${\mathop{\rm t}\nolimits} {\rm{anx}} + \cot 2x = \frac{{\sin {\rm{x}}\sin 2x + \cos x\cos
2x}}{{\cos x\sin 2x}} = \frac{{\cos x}}{{\cos x\sin 2x}}$$ = \frac{1}{{\sin 2x}}$
$
\Rightarrow f(x) = \sin 3x\sin 4x\sin 2x = \frac{1}{2}\left( {\cos x - c{\rm{os}}7x} \right)\sin 2x $$
= \frac{1}{2}\sin 2x\cos x - \frac{1}{2}\sin 2x\cos 7x\\ $
$= \frac{1}{4}\left( {{\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 3x} \right) - \frac{1}{4}\left( {\sin
( - 5x) + \sin 9x} \right) $
$= \frac{1}{4}{\mathop{\rm s}\nolimits} {\rm{inx}} +
\frac{1}{4}{\mathop{\rm s}\nolimits} {\rm{in3x + }}\frac{1}{4}{\mathop{\rm s}\nolimits}
{\rm{in5x - }}\frac{1}{4}{\mathop{\rm s}\nolimits} {\rm{in9x}}\\
\Rightarrow \int {f(x)dx} = - \frac{1}{4}\cos x - \frac{1}{{12}}c{\rm{os}}3x -
\frac{1}{{20}}c{\rm{os}}5x + \frac{1}{{36}}c{\rm{os}}9x + C
$
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