作者LuisSantos (^______^)
看板trans_math
标题Re: [积分]
时间Wed Apr 11 00:22:39 2007
※ 引述《hushenpaul (适之)》之铭言：
: ∫e^-x cosx dx=?
令 u = e^(-x) , dv = cosx dx
则 du = -e^(-x) dx , v = sinx
∫(e^(-x))(cosx) dx
= (e^(-x))(sinx) - ∫(sinx)(-e^(-x)) dx
= (e^(-x))(sinx) + ∫(e^(-x))(sinx) dx
= (e^(-x))(sinx) + (e^(-x))(-cosx) - ∫(-cosx)(-(e^(-x))) dx
(令 u = e^(-x) , dv = sinx , 则 du = -e^(-x) dx , v = -cosx)
= (e^(-x))(sinx) - (e^(-x))(cosx) - ∫(e^(-x))(cosx) dx
(2)(∫(e^(-x))(cosx) dx) = (e^(-x))(sinx) - (e^(-x))(cosx)
1
∫(e^(-x))(cosx) dx = (---)((e^(-x))(sinx) - (e^(-x))(cosx)) + c
2
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