Question #5c72a

Question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-30T02:03:29

#int (sin^8x-cos^8x)/(1-2sin^2xcos^2x)dx#

#=int ((sin^4x+cos^4x)(sin^4x-cos^4x))/((sin^2x+cos^2x)^2-2sin^2xcos^2x)dx#

#=int ((sin^4x+cos^4x)(sin^4x-cos^4x))/(sin^4x+cos^4x)dx#

#=int (sin^4x-cos^4x)dx#

#=int (sin^2x-cos^2x)(sin^2x+cos^2x)dx#

#=int (sin^2x-cos^2x)dx#

#=-intcos2xdx#

#=-1/2sin2x#