Integrate the following function: 
Evaluate:
  $$\int \frac{1}{ \cos^4x+ \sin^4x}dx$$

Tried making numerator $\sin^2x+\cos^2x$
making numerator $(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
Dividing throughout by $cos^4x$
Thank you in advance

Let me try something mixing the work of chndn and lab bhattacharjee.  First let's rewrite the denominator as chndn has tried.
$$\int\dfrac1{\cos^4x+\sin^4x}dx=\int\dfrac1{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x}dx=$$
$$\int\dfrac1{1-\frac12\sin^22x}dx$$
Now let's do a move similar to lab bhattacharjee's.
$$\int\dfrac{\csc^22xdx}{\csc^22x-\frac12}=\int\dfrac{\csc^22xdx}{\cot^22x+\frac12}$$
Now let $u=\cot 2x$.
