∫dx/(1+2cosx)

网友回答

cosx = 1 - 2sin²(x/2)
2cosx = 2 - 4sin²(x/2)
∫ dx/(1 + 2cosx)
= ∫ dx/[3 - 4sin²(x/2)]
= ∫ dx/[3sin²(x/2) + 3cos²(x/2) - 4sin²(x/2)]
= ∫ dx/[3cos²(x/2) - sin²(x/2)]
= ∫ sec²(x/2)/[3 - tan²(x/2)] dx
= 2∫ 1/{[√3 - tan(x/2)][√3 + tan(x/2)]} dtan(x/2)
= 2/(2√3)∫ { [√3 + tan(x/2)] + [√3 - tan(x/2)] }/{ [√3 - tan(x/2)][√3 + tan(x/2)] } dtan(x/2)
= (1/√3)∫ {1/[√3 - tan(x/2) + 1/[√3 + tan(x/2)] } dtan(x/2)
= (1/√3) { - ln|√3 - tan(x/2)| + ln|√3 + tan(x/2)| } + C
= (1/√3)ln| [√3 + tan(x/2)]/[√3 - tan(x/2)] | + C
或使用万能代换：令u = tan(x/2),dx = 2du/(1 + u²),cosx = (1 - u²)/(1 + u²)
∫ dx/(1 + 2cosx)
= ∫ 1/[1 + 2(1 - u²)/(1 + u²)] • 2/(1 + u²) • du= 2∫ du/(3 - u²) = 2∫ du/[(√3 - u)(√3 + u)]
= (2/2√3)∫ [(√3 + u) + (√3 - u)]/[(√3 - u)(√3 + u)] du
= (1/√3)[- ln|√3 - u| + ln|√3 + u|] + C
= (1/√3)ln| (√3 + u)/(√3 - u) | + C
= (1/√3)ln| [√3 + tan(x/2)]/[√3 - tan(x/2)] | + C
,实际上跟上面的步骤一样,但用u替代了tan(x/2)