∫lncosxdx(o≤x≤π/2)求解.x=π/2为瑕点，求大虾……

Question

∫lncosx dx(o≤x≤π/2)求解.
x=π/2为瑕点，求大虾……

Answer 1

I = ∫ [0,π/2] lncosx dx = ∫ [0,π/2] lnsinx dx
= ∫ [0,π/4] lncosx dx + ∫ [π/4,π/2] lncosx dx
利用下式：
∫ [0,π/4] lnsinx dx
= ∫ [0,π/4] lncos(π/2 -x) dx 令 u = π/2 -x
= ∫ [π/2,π/4] lncosu (-1)du
= ∫ [π/4,π/2] lncosu du
I = ∫ [0,π/4] lncosx dx + ∫ [0,π/4] lnsinx dx = ∫ [0,π/4] 【lncosx + lnsinx】 dx
= ∫ [0,π/4] ln (1/2) sin2x dx
= - π ln2 / 4 + (1/2) ∫ [0,π/2] lnsinu du 令 u = 2x
= - π ln2 / 4 + (1/2) I
=》 I = - π ln2 / 2