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x cotx dcotx的不定积分
人气:449 ℃ 时间:2020-06-30 16:05:04
解答
∫x*cotx dcotx
= (1/2)∫x d(cot²x)
= (1/2)xcot²x - (1/2)∫cot²x dx
= (1/2)xcot²x - (1/2)∫(csc²x-1) dx
= (1/2)x(csc²x-1) - (1/2)(-cotx-x) + C
= (1/2)xcsc²x - x/2 + (1/2)cotx + x/2 + C
= (1/2)(xcsc²x + cotx) + C
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