Question #5c72a

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Answer 1 · 2017-06-30T02:03:29

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

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

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

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

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

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

$=-intcos2xdx$

$=-1/2sin2x$