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不定积分∫(xcosx)/sin²x dx等于多少?
人气:141 ℃ 时间:2019-09-24 05:05:58
解答
∫[xcosx/(sinx)^2]dx
=∫[x/(sinx)^2]d(sinx)
=-∫xd(1/sinx)
=-x/sinx+∫(1/sinx)dx
=-x/sinx+∫[sinx/(sinx)^2]dx
=-x/sinx-∫[1/(sinx)^2]d(cosx)
=-x/sinx-(1/2)∫{(1+cosx+1-cosx)/[(1+cosx)(1-cosx)]}d(cosx)
=-x/sinx-(1/2)∫[1/(1-cosx)]d(cosx)-(1/2)∫[1/(1+cosx)]d(cosx)
=-x/sinx+(1/2)ln(1-cosx)-(1/2)ln(1+cosx)+C
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