Question

高等数学 第四题

Answer 1

4. (x-1)^2+y^2 = 4, y ≥ 0, 即 x = 1+2cost, y = 2sint, 0 ≤ t ≤ π., 则
I = ∫<0, π> [1+2(cost-sint)](-2sint) + [1+2(cost+sint)](2cost)]dt/[(1+2cost)^2+4(sint)^2]
= ∫<0, π> [4+2(cost-sint)]dt/(5+4cost)
= ∫<0, π>[(5/2+2cost)+3/2]dt/(5+4cost) + (1/2)∫<0, π>d(5+4cost)/(5+4cost)
= ∫<0, π>(1/2)dt + (3/2)∫<0, π>dt/(5+4cost) + (1/2)[ln(5+4cost)]<0, π>
= π/2 - (1/2)ln5 + (3/2)∫<0, π>dt/(5+4cost) , 后者令 u = tan(t/2), 则
I = π/2 - (1/2)ln5 + 3 ∫<0, +∞>du/(9+u^2)
= π/2 - (1/2)ln5 + [arctan(u/3)] <0, +∞>
= π/2 - (1/2)ln5 + π/2 = π - (1/2)ln5