integral $\int \frac{1}{sin^{6}x+cos^{6}x} $ Hi I have checked my working out for numerous times but I could not find any error in it. Could you please have a look at my working out and point out my mistake. Underneath the working out is the correct answer.
Note

$$\tan^{-1} \frac{\tan x}{1-\tan^2x}
 =\tan^{-1} \frac12\tan 2x 
=\tan^{-1} \frac1{2\cot 2x }\\
= \frac\pi2 - \cot^{-1}  \frac1{2\cot 2x }
= \frac\pi2 - \tan^{-1} (2\cot 2x ) $$

which is equivalent to the answer provided.
For $\tan x,\cot x\ne0,$

$$\cot x-\tan x=\dfrac{1-\tan^2x}{\tan x}=2\cot2x$$

Alternatively,

$$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$$

$$\dfrac1{\sin^6x+\cos^6x}=\dfrac1{1-3\sin^2x\cos^2x}=\dfrac4{4-3\sin^22x}=\dfrac{4\csc^22x}{4(1+\cot^22x)-3}$$

Now set $\cot2x=y$