<s>[INST] Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ 
Problem: Find  $\displaystyle\int \frac {\tan 2x} {\sqrt {\cos^6 x +\sin^6 x}} dx $
Solution: $\tan 2x= \dfrac{2\tan x}{1-\tan^2 x}$
Also I can take $\cos^6x$ common from $\sqrt {\cos^6x +\sin^6x}$

I don't know whether it is good approach to the question
Please help
 [/INST] HINT:
$$\cos^6x+\sin^6x=(\cos^2x+\sin^2x)^3-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$
$$=1-3\cos^2x\sin^2x=1-\frac34(\sin2x)^2$$
$$=1-3\cos^2x\sin^2x=1-
\frac34(\sin2x)^2=1-\frac34(1-\cos^22x)=\frac{1+3\cos^22x}4$$
Use $\cos2x=u$
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