Question

∫dx/(a+bcosx)

Answer 1

当 b = a 时，I = ∫dx/(a+bcosx) = (1/a)∫dx/(1+cosx)
= (1/a)∫d(x/2)/[cos(x/2)]^2 = (1/a)tan(x/2) + C;
当 b ≠ a 时，用半角代换，令 tan(x/2) = u, 则 x = 2arctanu,

dx = 2du/(1+u^2), cosx = (1-u^2)/(1+u^2)
I = ∫dx/(a+bcosx) = ∫2du/[a(1+u^2)+b(1-u^2)] = 2∫du/[(a+b)+(a-b)u^2]
a > b 且 a+b > 0 时， I = [2/√(a^2-b^2)]arctan[tan(x/2)/k] + C
其中 k = √[(a+b)/(a-b)];
另有 a > b 且 a+b < 0 ; a < b, a+b < 0 ; a < b, a+b > 0 分别讨论。

Answer 2

就万能代换