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

Another way:
$$I=\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{(1+\tan^2x)\sec^2x dx}{1+\tan^4x}$$
Setting $\displaystyle\tan x=u,$
$$I=\frac{(1+u^2)du}{1+u^4}=\int\frac{1+\dfrac1{u^2}}{u^2+\dfrac1{u^2}}du$$
$$=\int\frac{1+\dfrac1{u^2}}{\left(u-\dfrac1u\right)^2+2}du$$
Set $\displaystyle u-\dfrac1u=v$
