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