Question

高等数学，题目如图。。

高等数学，题目如图。。求详细步骤，满意必采纳，谢谢

Answer 1

化极坐标
I = ∫<0, π/2> dt ∫<0, 1> √[(1-r^2)/(1+r^2)] rdr
= (1/2)∫<0, π/2> dt ∫<0, 1> √[(1-r^2)/(1+r^2)] dr^2 (令 r^2 = u)
= (π/4)∫<0, 1> √[(1-u)/(1+u)] du
令 √[(1-u)/(1+u)] = v, 则 1-u = v^2(1+u), u = (1-v^2)/(1+v^2),
du = [-2v(1+v^2)-2v(1-v^2)]dv/(1+v^2)^2 = -4vdv/(1+v^2)^2.
u = 0 时， v = 1； u = 1 时， v = 0
I = (π/4)∫<1, 0> v[-4vdv/(1+v^2)^2] = π∫<0, 1> v^2dv/(1+v^2)^2
= π∫<0, 1> (1+v^2-1)dv/(1+v^2)^2
= π∫<0, 1> dv/(1+v^2) - π∫<0, 1> dv/(1+v^2)^2
后者令 v = tant， 则
π∫<0, 1> dv/(1+v^2)^2 = π∫<0, π/4> (sect)^2dt/(sect)^4
= π∫<0, π/4>(cost)^2dt = (π/2)∫<0, π/4>(1+cos2t)dt
= (π/2)[t+(1/2)sin2t]<0, π/4> = (π/2)(π/4 + 1/2)
则 I = π[arctanv]<0, 1> - (π/2)(π/4 + 1/2)
= π^2/4 - π^2/8 - π/4 = π^2/8 - π/4