Question

求解 定积分问题！

Answer 1

不妨设 a > 0，
当 a = 1 时， I = ∫<0, π/2>2dx/(sinx)^2 = -[cotx]<0, π/2> = +∞.
当 a ≠ 1 时，I = ∫<0, π/2>2dx/[1-(acosx)^2]
= ∫<0, π/2>dx/(1-acosx) + ∫<0, π/2>dx/(1+acosx), 令 u = tan(x/2)
= ∫<0, 1>2du/[(1-a)+(1+a)u^2] + ∫<0, 1>2du/[(1+a)+(1-a)u^2]
当 0 < a < 1 时， 记 b^2 = (1-a)/(1+a)
I = [2/(1+a)][(1/b)arctan(x/b)]<0, 1> + [2/(1-a)][barctan(bx)]<0, 1>
= {2/[b(1+a)]}arctan(1/b) + [2b/(1-a)]arctanb
当 a > 1 时，记 b^2 = (a-1)/(1+a)
I = [1/(1+a)](1/b)ln|(x-b)/(x+b)|]<0, 1> + [b/(a-1)][ln|(1/b+x)/(1/b-x)|]<0, 1>
= {1/[b(1+a)]}ln[(1-b)/(1+b)] + [b/(a-1)]ln[(1+b/(1-b)]