18/06/16 01:23:30.75 Sq4cRvDq.net
>>140
t = tan(x/2) とおくと
∫ cos(x)/{1+cos(x)} dx = ∫{1 - 1/[1+cos(x)]} dx
= ∫ {1 - 1/[2cos(x/2)^2]} dx
= x - ∫ 1/[cos(x/2)]^2 d(x/2)
= x - tan(x/2) +c
= x - sin(x)/[1+cos(x)] + c,,
18/06/16 01:23:30.75 Sq4cRvDq.net
>>140
t = tan(x/2) とおくと
∫ cos(x)/{1+cos(x)} dx = ∫{1 - 1/[1+cos(x)]} dx
= ∫ {1 - 1/[2cos(x/2)^2]} dx
= x - ∫ 1/[cos(x/2)]^2 d(x/2)
= x - tan(x/2) +c
= x - sin(x)/[1+cos(x)] + c,,