∫0π/2(sinx-cosx)/(1+sin2x)dx.

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问题0π/2(sinx-cosx)/(1+sin2x)dx.

选项

答案0π/2(sinx-cosx)/(1+sin2x)dx=-∫0π/2d(sinx+cosx)/(sinx+cosx)2=1/(sinx+cosx)|0π/2=0.

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