求积分(1+sinx)/[sinx*(1+cosx)]dx
求积分(1+sinx)/[sinx*(1+cosx)]dx
数学人气:642 ℃时间:2019-11-12 23:48:04
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∫{(1+sinx)/[sinx(1+cosx)]}dx=∫{1/[sinx(1+cosx)]}dx+∫[1/(1+cosx)]dx=∫{sinx/[(six)^2(1+cosx)]}dx+(1/2)∫{[1/[cos(x/2)]^2}dx=-∫{1/[(1-cosx)(1+cos...看着眼晕 还是谢谢你 不过∫{1/[cos(x/2)]^2}d(x/2)=∫[sec(x/2)]^2d(x/2)=tan(x/2)+C这里是你错了还是我错了抱歉!是我们都错了。∫{(1+sinx)/[sinx(1+cosx)]}dx=∫{1/[sinx(1+cosx)]}dx+∫[1/(1+cosx)]dx=∫{sinx/[(six)^2(1+cosx)]}dx+(1/2)∫{[1/[cos(x/2)]^2}dx=-∫{1/[(1-cosx)(1+cosx)^2]}d(cosx)+∫{1/[cos(x/2)]^2}d(x/2)=-(1/2)∫{(1-cosx+1+cosx)/[(1-cosx)(1+cosx)^2]}d(cosx) +tan(x/2)=-(1/2)∫[1/(1+cosx)^2]d(cosx) -(1/2)∫{1/[(1-cosx)(1+cosx)]d(cosx)+tan(x/2)=(1/2)[1/(1+cosx)]+tan(x/2) -(1/4)∫{(1-cosx+1+cosx)/[(1-cosx)(1+cosx)]}d(cosx)=1/(2+2cosx)+tan(x/2)-(1/4)∫[1/(1+cosx)]d(cosx) -(1/4)∫[1/(1-cosx)]d(cosx)=1/(2+2cosx)+tan(x/2)-(1/4)ln(1+cosx)+(1/4)ln(1-cosx)+C。
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