∫（3-sinx）/(3+cosx)dx怎么解,请给位大侠帮帮忙

Question

∫（3-sinx）/(3+cosx)dx怎么解,请给位大侠帮帮忙

Answer 1

令 3 - sinx = A(3 + cosx) + B(3 + cosx)' + C
3 - sinx = A(3 + cosx) + B(- sinx) + C
3 - sinx = - Bsinx + Acosx + (3A + C)
- 1 = - B => B = 1
A = 0
3 = 3A + C => C = 3
∫ (3 - sinx)/(3 + cosx) dx
= ∫ (3 + cosx)'/(3 + cosx) dx + 3∫ dx/(3 + cosx)
= ln(3 + cosx) + 3(1/2)∫ dx/[1 + (1 + cosx)/2]
= ln(3 + cosx) + (3/2)∫ dx/[sin²(x/2) + 2cos²(x/2)]
= ln(3 + cosx) + 3∫ sec²(x/2)/[2 + tan²(x/2)] d(x/2)
= ln(3 + cosx) + 3∫ d[tan(x/2)]/[2 + tan²(x/2)]
= ln(3 + cosx) + (3/√2)arctan[(1/√2)tan(x/2)] + C
对于∫ dx/(3 + cosx),用换元法u = tan(x/2)也可.