Question

求∫cos(ln(1/(1-x)))dx

Answer 1

I = ∫cos{ln[1/(1-x)]}dx = ∫cos{-ln(1-x)}dx = ∫cos[ln(1-x)]dx
令 ln(1-x) = u， 则 1-x = e^u, x = 1- e^u, dx = -e^udu,
I = -∫cosue^udu = -∫cosude^u = -e^ucosu + ∫e^udcosu
= -e^ucosu - ∫e^usinudu = -e^ucosu - ∫sinude^u
= -e^ucosu - e^usinu + ∫e^ucosudu = -e^u(cosu+sinu) - I
I = -(1/2)e^u(cosu+sinu) + C
= -(1/2)e^[ln(1-x)]{(cos[ln(1-x)]+sin[ln(1-x)]} + C
= -(1/2)(1-x){cos[ln(1-x)] + sin[ln(1-x)]} + C

Answer 2

把-1提到ln之外，并令t等于ln（1-x），dx等于-d(1-x)，则被积函数为-(cost)(et),再用分部积分法计算即可