※ 引述《freshcute (咕嚕咕嚕)》之銘言：
: ∫cos(lnx)dx
: 請問要怎麼解呢0.0
令 y = lnx , 則 x = e^y => dx = e^y dy
∫cos(lnx) dx
= ∫(cos(y))(e^y) dy
= ∫(e^y)(cos(y)) dy
= (e^y)(sin(y)) - ∫(e^y)(sin(y)) dy
(令 u = e^y , dv = cos(y) dy , 則 du = e^y dy , v = sin(y))
= (e^y)(sin(y)) - ((e^y)(-cos(y)) - ∫(e^y)(-cos(y)) dy)
(令 u = e^y , dv = sin(y) dy , 則 du = e^y dy , v = -cos(y))
= (e^y)(sin(y)) + (e^y)(cos(y)) - ∫(cos(y))(e^y) dy
(2)(∫(cos(y))(e^y) dy) = (e^y)(sin(y)) + (e^y)(cos(y))
(e^y)(sin(y)) + (e^y)(cos(y))
∫(cos(y))(e^y) dy = ----------------------------- + c
2
(x)(sin(lnx)) + (x)(cos(lnx))
∫cos(lnx) dx = ----------------------------- + c
2
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