高数 求不定积分 有详解更好哇!
2个回答
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因为
1/(1+cosx)
=1/2*1/(cos^2(x/2)
=sec^2(x/2)/2
sinx/(1+cosx)
=2*sin(x/2)*cos(x/2)/(2*cos^2(x/2)]
=tan(x/2)
原式=∫e^x/(1+cosx)dx+∫e^xsinx/(1+cosx)dx
=∫e^x*sec^2(x/2)/2dx+∫e^x*tan(x/2)dx
=∫e^xd(tan(x/2))+∫tan(x/2)d(e^x)
=e^xtan(x/2)-∫tan(x/2)d(e^x)+∫tan(x/2)d(e^x) (这一步是用分布积分法,把第一个积分拆了)
=e^xtan(x/2)+C
1/(1+cosx)
=1/2*1/(cos^2(x/2)
=sec^2(x/2)/2
sinx/(1+cosx)
=2*sin(x/2)*cos(x/2)/(2*cos^2(x/2)]
=tan(x/2)
原式=∫e^x/(1+cosx)dx+∫e^xsinx/(1+cosx)dx
=∫e^x*sec^2(x/2)/2dx+∫e^x*tan(x/2)dx
=∫e^xd(tan(x/2))+∫tan(x/2)d(e^x)
=e^xtan(x/2)-∫tan(x/2)d(e^x)+∫tan(x/2)d(e^x) (这一步是用分布积分法,把第一个积分拆了)
=e^xtan(x/2)+C
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