$\int \frac{1}{\sin ^6 x+\cos ^6 x} d x$ is equal to

$\tan ^{-1}(\tan x-\cot x)+C$

Let

$I =\int \frac{1}{\sin ^6 x+\cos ^6 x} d x=\int \frac{\sec ^6 x}{1+\tan ^6 x} d x$

$\Rightarrow I =\int \frac{\left(1+\tan ^2 x\right)^2}{1+\tan ^6 x} \sec ^2 x d x$

$=\int \frac{\left(1+t^2\right)^2}{1+t^6} d t$, where $t=\tan x$

$\Rightarrow I=\int \frac{t^2+1}{t^4-t^2+1} d t=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-1} d t=\int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+1^2}$

$\Rightarrow  I=\tan ^{-1}\left(t-\frac{1}{t}\right)+C=\tan ^{-1}(\tan x-\cot x)+C$