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Ta có
$\dfrac{\sin x\cos x}{4\sin^{2} x+7\cos^{2} x+3\cos 2x}=\dfrac{2\sin x\cos x}{8\sin^{2} x+14\cos^{2} x+6\cos 2x}=\dfrac{\sin2 x}{4(1-\cos2x)+7(1+\cos2x)+6\cos 2x}$
$=\dfrac{\sin2 x}{9\cos 2x+11}=-\dfrac{1}{18}.\dfrac{(9\cos 2x+11)'}{9\cos 2x+11}$
Suy ra
$\int\limits_{}^{} \dfrac{\sin x\cos x}{4\sin^{2} x+7\cos^{2} x+3\cos 2x}dx=-\dfrac{1}{18}\ln|9\cos 2x+11|+C$
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