21/03/17 00:14:48.43 Rkkg81B/.net
>>910
厳密解
y = log(1+cos(x)),
y ' = - sin(x)/(1+cos(x)),
√{1+(y ')^2} = √{2/(1+cos(x))} = 1/cos(x/2),
より
L = ∫√{1+(y ')^2} = ∫1/cos(x/2) dx
= log|(1+sin(x/2)/(1-sin(x/2))|
= - 2log|tan((π-x)/4))|,
tan(π/4) = 1, tan(π/8) = √2 - 1.