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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

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  • $\begingroup$ Simplifying the surd in the denominator might help $\endgroup$ Jul 17 '13 at 5:30
  • $\begingroup$ 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). $\endgroup$ Jul 17 '13 at 5:36
  • $\begingroup$ @ZevChonoles,thanks,in future I will take care of it $\endgroup$
    – rst
    Jul 17 '13 at 5:41
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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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  • $\begingroup$ Thanks,now it is easy to solve $\endgroup$
    – rst
    Jul 17 '13 at 5:39
  • $\begingroup$ @rst, my pleasure $\endgroup$ Jul 17 '13 at 5:44

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