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

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

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

If $\tan x=p$, then $\sec ^2 x d x=d p$

$=\frac{\left(1+p^2\right)^2 d p}{1+p^6}=\int \frac{\left(1+p^2\right)}{p^4-p^2+1} d p=\int \frac{p^2\left(1+\frac{1}{p^2}\right)}{p^2\left(p^2+\frac{1}{p^2}-1\right)} d p $

∴ $I=\int \frac{d k}{k^2+1}=\tan ^{-1}(k)+c$

If $p-\frac{1}{p}=k$, then $\left(1+\frac{1}{p^2}\right) d p=d k$

$=\tan ^{-1} \left(p-\frac{1}{p}\right)+c=\tan ^{-1} (\tan x-\cot x)+c=\tan ^{-1} (-2 \cot 2 x)+c$

Hence (3) is the correct answers.