看起來很好積分但實際上卻不能積分的函數，arccosx除cosx奇函數(ft.

[@鯉魚王-e5z]

封面左邊是侯友宜嗎

[@沈博智-x5y]

我的方法
積分從 -pi/4 到 pi/4 arccos(x)/cos(x) dx
= 積分從 -pi/4 到 pi/4 (pi/2 - arcsin(x))/cos(x) dx
= 積分從 -pi/4 到 pi/4 [(pi/2)/cos(x) - arcsin(x)/cos(x)] dx
= 積分從 -pi/4 到 pi/4 (pi/2)/cos(x) dx - 積分從 -pi/4 到 pi/4 arcsin(x)/cos(x) dx
arcsin(x) 是奇的
分母 cos(x) 是偶的
所以 arcsin(x)/cos(x) 是奇的
=> 積分從 -pi/4 到 pi/4 arcsin(x)/cos(x) dx = 0
積分從 -pi/4 到 pi/4 arccos(x)/cos(x) dx
= pi/2 積分從 -pi/4 到 pi/4 1/cos(x) dx
= (pi/2)(ln|sec(pi/4) + tan(pi/4)| - ln|sec(-pi/4) + tan(-pi/4)|)
= (pi/2)(ln|根號 2 + 1| - ln|2^(1/2) - 1|)
= (pi/2)ln|(sqrt(2)+1)/(sqrt(2)-1)|
= (pi/2)ln|((sqrt(2)+1)^2)/(2-1)|
= (pi/2)ln|(sqrt(2) + 1)^2|
= (pi/2)ln((sqrt(2) + 1)^2)
= pi ln|sqrt2 + 1|
= pi ln(1 + sqrt(2))
= pi ln(1 + sqrt(1 + 1^2))
= pi arsinh(1)
edit: I see bprp basically did the same method in original livestream.

[@沈博智-x5y]

其實 equivalent 答案
(pi/2)ln((sqrt(2)+1)/sqrt(2)-1) = pi arcoth(sqrt(2))
(pi/2)ln((sqrt(2)+1)/sqrt(2)-1) = (pi/2)ln((1 + 1/sqrt(2))(1 - 1/sqrt(2))) = pi artanh(1/sqrt(2))
pi ln(1+sqrt(2)) = pi arsinh(1) = pi artanh(1/sqrt(2)) = pi arcoth(sqrt(2))