令:x=π/2-t∫[0,π/2] 1/√(1-sinx) dx =∫[π/2,0] 1/√(1-cost) (-dt) =∫[0,π/2] 1/√(1-cost) dt∵ lim(t->0+) [1/√(1-cost)]/(1/t) = lim(t->0+) √[t^2/(1-cost)]= √2及:∫[0,π/2] 1/t dt 发散,由瑕积分...