If $\int \frac{\mathrm{d}x}{{\cos }^{6}x+{\sin }^{6}x}={\tan }^{-1}(k\tan 2x)+C$, then

$\begin{array}{l}\frac{1}{{\cos }^{6}x+{\sin }^{6}x}=\frac{1}{1-3{\sin }^{2}x{\cos }^{2}x}\\ \frac{{\sec }^{4}x}{{\sec }^{4}x-3{\tan }^{2}x}=\frac{\left(1+{\tan }^{2}x\right){\sec }^{2}x}{{\left(1+{\tan }^{2}x\right)}^2-3{\tan }^{2}x}\end{array}$

Putting $\tan x=t$ the given integral reduces to

$\begin{array}{l}\int \frac{1+t^2}{1+t^4-t^2}dt=\int \frac{1+1/t^2}{t^2+1/t^2-1}dt\\ =\int \frac{du}{u^2+1},u=t-1/t\\ ={\tan }^{-1}(\tan x-\cot x)+C\\ ={\cot }^{-1}\left(-\frac{1}{2}\tan 2x\right)+C\\ ={\tan }^{-1}\left(\frac{1}{2}\tan 2x\right)+\text{ Const. }\end{array}$