∫sin2x/sin^4x+cos^4xdx?

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∫sin2x/sin^4x+cos^4xdx?

∫sin2x/sin^4x+cos^4xdx?
∫sin2x/sin^4x+cos^4xdx?

∫sin2x/sin^4x+cos^4xdx?
∫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
=-arctancos2x+c

题目是∫sin2x/(sin^4x+cos^4x)dx 这样的吗。晕