# 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

• Simplifying the surd in the denominator might help Jul 17 '13 at 5:30
• Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. Some MathJax advice: Named math operators should appear upright, and the common ones have their own code for this purpose (e.g. \sin, \log - see entry 11 in our MathJax guide). Jul 17 '13 at 5:36
• @ZevChonoles,thanks,in future I will take care of it
– rst
Jul 17 '13 at 5:41

$$\cos^6x+\sin^6x=(\cos^2x+\sin^2x)^3-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$ $$=1-3\cos^2x\sin^2x=1-\frac34(\sin2x)^2$$ $$=1-3\cos^2x\sin^2x=1- \frac34(\sin2x)^2=1-\frac34(1-\cos^22x)=\frac{1+3\cos^22x}4$$
Use $\cos2x=u$