trigonometric integration

d/dx ( sin^4/4 - sin^6/6 + C) =

sin^3x(cosx) - sin^5x(cosx) =

cosx(sin^3x)(1 - sin^2x) =

cosx (sin^3x)(cos^2x) =

sin^3xcos^3x

And           

d/dx(-cos^4x/4 + cos^6x/6 + C)=

-(sinx)(-cos^3x)   - (sinx)cos^5x  =

(sinx)(cos^3x)(1 - cos^2x) =

(sinx)(cos^3x)(sin^2x) =

(cos^3x)(sin^3x)  = sin^3xcos^3x

And they are exactly the same.....!!!

Thank you CPhill