∫sin2x/sin^4x+cos^4xdx=∫sin2x/[(sin²x+cos²x)²-2sin²xcos²x]dx=∫sin2x/[1-1/2sin²2x]dx=2∫sin2x/[2-sin²2x]dx=2∫sin2x/(1+cos²2x)dx=-∫1/(1+cos²2x)dcos2x=-a...