#int(sin^6x+cos^6x)/(sin^2xcos^2x)dx#
Firstly, we need to simplify the integral to make it more accessible
So let
#" "s=sinx, c=cosx#
#=int((s^6+c^6)/(s^2c^2))dx#
Rewriting and simplifying the integral
#(s^6+c^6)/(s^2c^2)=((s^3)^2+(c^3)^2)/(s^2c^2)#
The numerator is sum of cubes
Using
#a^3+b^3=(a+b)(a^2-ab+b^2)#
We have,
#((s^2)^3+(c^2)^3)/(s^2c^2)=((s^2+c^2)(s^4-s^2c^2+c^4))/(s^2c^2)#
But #s^2+c^2=1#
#:.=((s^4-s^2c^2+c^4))/(s^2c^2)#
Now substitute back the sines and cosines
#=(sin^4x-sin^2xcos^2x+cos^4x)/(sin^2xcos^2x)#
#=sin^4x/(sin^2xcos^2x)-(sin^2xcos^2x)/(sin^2xcos^2x)+cos^4x/(sin^2xcos^2x)#
#=sin^2x/cos^2x-1+cos^2x/sin^2x#
#=tan^2x-1+cot^2x#
now back to the integral!
#int(sin^6x+cos^6x)/(sin^2xcos^2x)dx#
#=int(tan^2x-1+cot^2x)dx#
#=int(sec^2x-1-1+csc^2x-1)dx#
#=int(sec^2x-3+csc^2x)dx#
These are standard integrals!!
#=tanx-3x-cotx+C#
