Integral

\hspace{-16}$Given $\bf{\int \frac{1}{\sin^ 6 x+\cos^6 x}dx}$\\\\\\ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{\cos^6 x}$\\\\\\ $\bf{ = \int \frac{\sec^6 x}{1+\tan^6 x}dx = \int \frac{(1+\tan^2 x)^2 \cdot \sec^2 x}{1+\tan^6 x}dx}$\\\\\\ Now Let $\bf{\tan x= t\;,}$ Then $\bf{\sec^2 xdx = dt}$\\\\\\ $\bf{=\int\frac{(1+t^2)^2}{1+t^6}dt = \int\frac{(1+t^2)^2}{(1+t^2)\cdot (1+t^4-t^2)}dt}$\\\\\\ Using $\bf{\bullet\;\; 1+(t^2)^3 = (1+t^2)\cdot (1+t^4-t^2)}$\\\\\\$\bf{=\int\frac{1+t^2}{1+t^4-t^2}dt}$\\\\\\ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{t^2}$\\\\\\ $\bf{=\int\frac{1+t^{-2}}{t^{-2}+t^{2}-1}dt = \int\frac{t^{-2}+1}{\left(t-t^{-1}\right)^2+1^2}dt}$\\\\\\ Now Let $\bf{\left(t-\frac{1}{t}\right)=u\;,}$ Then $\bf{\left(1+\frac{1}{t^2}\right)dt = du}$\\\\\\ $\bf{=\int\frac{1}{1+u^2}du = \tan^{-1}(u)+\mathbb{C} = \tan^{-1}\left(\frac{t^2-1}{t}\right)+\mathbb{C}}$\\\\\\

\hspace{-16}$So $\bf{\int\frac{1}{\sin^6 x+\cos^6 x}dx = \tan^{-1}\left(\frac{\tan^2 x-1}{\tan x}\right)+\mathbb{C}}$