\[\begin{array}{l}
\dfrac{1}{{\tan x + \cot 2x}} = \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos 2x}}{{\sin 2x}}}} = \dfrac{1}{{\dfrac{{\sin 2x\sin x + \cos 2x\cos x}}{{\cos x\sin 2x}}}} = \dfrac{{\cos x\sin 2x}}{{\cos (2x - x)}} = \sin 2x\\
I = \int {\dfrac{{\sin 3x\sin 4xdx}}{{\tan x + \cot 2x}} = } \int {\sin 3x.2{{\sin }^2}2x\cos 2xdx} = \int {\sin 3x(1 - \cos 4x)\cos 2xdx} \\
I = \int {\sin 3x\cos 2xdx} - \int {\sin 3x\cos 4x\cos 2xdx} \\
{I_1} = \int {\sin 3x\cos 2xdx} = \int {\dfrac{1}{2}(\sin x + \sin 5x)dx} = - \dfrac{{\cos x}}{2} - \dfrac{{\cos 5x}}{{10}} + {C_1}\\
{I_2} = \int {\sin 3x\cos 4x\cos 2xdx} = \int {\dfrac{1}{2}(\sin x\cos 2x + \sin 5x\cos 2x)dx} \\
{I_2} = \int {\dfrac{1}{4}(\sin 3x - \sin x + \sin 3x + \sin 7x)dx} = - \left( {\dfrac{{\cos 3x}}{6} - \dfrac{{\cos x}}{4} + \dfrac{{\cos 7x}}{{28}}} \right) + {C_2}\\
I = {I_1} - {I_2} = \dfrac{{\cos 7x}}{{28}} - \dfrac{{\cos 5x}}{{10}} + \dfrac{{\cos 3x}}{6} - \dfrac{{3\cos x}}{4} + C
\end{array}\]
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