$I=\int_{0}^{\frac{\pi }{6}}\frac{\tan^4x}{\cos 2x}\ dx$
$I=\int_{0}^{\frac{\pi }{6}}\frac{\tan^4x-1}{\cos 2x}\ dx+\int_{0}^{\frac{\pi }{6}}\frac{1}{\cos 2x}\ dx$
Xét $I_{1}=\int_{0}^{\frac{\pi }{6}}\frac{1}{\cos 2x}\ dx=\frac{1}{2}\int_{0}^{\frac{\pi }{6}}\frac{2\cos 2x}{1-\sin^{2}2x}\ dx$
Đặt $a=\sin 2x\Rightarrow da=2\cos 2x\ dx$
$\Rightarrow I_{1}= \frac{1}{4}\int_{0}^{\frac{\sqrt{3}}{2}}(\frac{1}{1-a}+\frac{1}{1+a})\ da=............................$
Xét $I_{2}=\int_{0}^{\frac{\pi }{6}}\frac{\tan^4x-1}{\cos 2x}\ dx$
$I_{2}=\int_{0}^{\frac{\pi }{6}}\frac{(\tan^2x-1)(\tan^2x+1)}{\cos 2x}\ dx$
$I_{2}=-\int_{0}^{\frac{\pi }{6}}(\tan^{2}x+1)^{2}\ dx$
Đặt $m=\tan x\Rightarrow dm=(1+tan^{2}x)\ dx$
$I_{2}=-\int_{0}^{\frac{\sqrt{3} }{3}}(m^{2}+1)\ dm=................................$
$cos2x=1-sin^22x$ à ?