令u = tan(x/2)，cosx = (1 - u²)/(1 + u²)，dx = 2du/(1 + u²)

∫ 1/(2 + cosx) * dx

= ∫ 1/[2 + (1 - u²)/(1 + u²)] * 2du/(1 + u²)

= ∫ (1 + u²)/(2 + 2u² + 1 - u²) * 2du/(1 + u²)

= 2∫ du/(u² + 3)，用公式：∫ dx/(x² + a²) = (1/a)arctan(x/a) + C，可得

= (2/√3)arctan(u/√3) + C

= (2/√3)arctan[(1/√3)tan(x/2)] + C，快很多吧

另解：

∫ dx/(2 + cosx)

= ∫ dx/[2sin²(x/2) + 2cos²(x/2) + cos²(x/2) - sin²(x/2)]，公式cos2x = cos²x - sin²x

= ∫ dx/[3cos²(x/2) + sin²(x/2)]，分子和分母再除以cos²(x/2)

= 2∫ sec²(x/2)/[3 + tan²(x/2)] d(x/2)

= 2∫ d[tan(x/2)]/[3 + tan²(x/2)]，湊sec²(x/2) d(x/2) = d[tan(x/2)]

= 2 * 1/√3 * arctan[tan(x/2)/√3] + C，公式：∫ dx/(x² + a²) = (1/a)arctan(x/a) + C

= (2/√3)arctan[(1/√3)tan(x/2)] + C

∫dx/(2+cosx)

cosx+1=2cos^2(x/2)

原式=∫dx/(2cos^2(x/2)+1)

=∫sec^(x/2)dx/(2+sec^2(x/2))

=2∫d(tg^(x/2))/(3+tg^2(x/2))

設tgx/2=t

=2∫dt/(1+(t/根號3)^2)

=2根號3∫d(t/根號3)/(1+(t/根號3)^2)

令t/根號3=tga

2根號3∫d(t/根號3)/(1+(t/根號3)^2)

=2根號3∫d(tga)/(1+(tga)^2)

=2根號3∫sec^2a/(sec^2a)da

=2根號3*a+c

a=arctg(t/根號3)

tg(x/2)=t

a=arctan(tg(x/2)/根號3))

原式=2根號（3）/3*arctan{（根號（3）/[3tan(x/2)]}

∫ sin³x/(2 + cosx) dx

= ∫ (cos²x - 1)/(2 + cosx) dcosx

= ∫ [cosx(cosx + 2 - 2) - 1]/(2 + cosx) dcosx

= ∫ [cosx(cosx + 2) - 2(cosx + 2 - 2) - 1]/(2 + cosx) dcosx

= ∫ [cosx(cosx + 2) - 2(cosx + 2) + 3]/(2 + cosx) dcosx

= ∫ [cosx - 2 + 3/(2 + cosx)] dcosx

= (1/2)cos²x - 2cosx + 3ln(2 + cosx) + c，這樣對吧？

用三角代換做，很容易的

就是用x=tan（t/2)代入，2代換成sin^2+cos^2