求不定积分∫[1/sin^2cos^2 (x)]dx
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求不定积分∫[1/sin^2cos^2(x)]dx求不定积分∫[1/sin^2cos^2(x)]dx求不定积分∫[1/sin^2cos^2(x)]dx∫[1/sin²xcos²x]
求不定积分∫[1/sin^2cos^2 (x)]dx
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求不定积分∫[1/sin^2cos^2 (x)]dx
∫[1/sin²xcos²x]dx
=∫[4/(2sinxcosx)²]dx
=∫[4/sin²2x]dx
=∫2csc²2xd2x
=-2∫(-csc²2x)d2x
=-2cot2x+C
根号下大于等于0
2x-4>=0,x>=2
4-2x>=0,x<=2
同时成立则x=2
所以2x-4=0,4-2x=0
所以y=0+0+x²=4
即x=2,y=4
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