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∫sinxcosx/(1+sin^4x)dx
人气:447 ℃ 时间:2020-06-04 20:57:12
解答
∫sinxcosx/(1+sin^4x)dx
=∫sinx/(1+sin^4x)d(sinx)
=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)
=1/2*arctan(sin^2x)+C
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