$∫\frac{dx}{cos^6x+sin^6x}$ is equal to

tan^-1 (-2 cot 2 x) + c

Let $I=∫\frac{dx}{cos^6x+sin^6x}=∫\frac{sec^6x}{1+tan^6x}dx=∫\frac{(1+tan^2x)^2sec^2xdx}{1+tan^6x}$

If tan x = p,  then sec² x dx = dp

$⇒I=∫\frac{(1+p^2)}{p^4-p^2+1}dp=∫\frac{p^2(1+\frac{1}{p^2})}{p^2(p^2+\frac{1}{p^2}-1)}dp$

 $=∫\frac{dk}{k^2+1}=tan^{-1}(k)+c(p-\frac{1}{p}=k,(1+\frac{1}{p^2})dp=dk)$

$=tan^{-1}(p-\frac{1}{p})+c=tan^{-1}(tanx-cotx)+c$

= tan^-1 (-2 cot 2 x) + c

Hence (C) and (D) are correct answers.