Ofabi MrThanh

\[\mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{4}} \frac{{dx}}{{{{\cos }^4}x.{{\sin }^2}x}}\]

$I = \int {\frac{{\sin x.\cos 2x}}{{\cos 5x}}dx}$

$$\dfrac{x+\sqrt{{x}^{2}+1}}{x-4}+\dfrac{2}{4{x}^{2}+1}=0$$

$$8\sqrt 2 \sin x.\cos 2x + 1 = \tan 4x + \tan x + \tan x.\tan 4x$$

MOD: Chú ý tiêu đề bạn nhé.