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

Let

    $\begin{array}{l}I=\int \frac{1}{{\sin }^{6}x+{\cos }^{6}x}dx=\int \frac{{\sec }^{6}x}{1+{\tan }^{6}x}dx\\ \Rightarrow I=\int \frac{{\left(1+{\tan }^{2}x\right)}^2}{1+{\tan }^{6}x}{\sec }^{2}xdx\end{array}$

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

     $\begin{array}{l}\Rightarrow I=\int \frac{t^2+1}{t^4}\frac{1}{-t^2+1}dt=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-1}dt=\int \frac{d\left(t-\frac{1}{t}\right)}{{\left(t-\frac{1}{t}\right)}^2+1^2}\\ \Rightarrow I={\tan }^{-1}\left(t-\frac{1}{t}\right)+C={\tan }^{-1}(\tan x-\cot x)+C\end{array}$