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∫(sinxconx / 1+(sin^4)x)dx
人气:429 ℃ 时间:2020-04-10 13:34:50
解答
∫(sinxconx /( (1+(sin²x)²)dx
=∫-sinx/( (1+(sin²x)²)dsinx
=-1/2∫1/( (1+(sin²x)²)dsin²x
=-1/2*arctan(sin²x)+C
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