大一高数定积分
展开全部
令x=sinθ,则:dx=cosθdθ,θ=arcsinx,
∴当x=1时,θ=π/2, 当x=1/√2时,θ=π/4。
∴原式=∫(π/4, π/2) √[1-(sinθ)^2]/(sinθ)^2 cosθdθ
=∫(π/4, π/2)[(cosθ)^2/(sinθ)^2]dθ
=∫(π/4, π/2) [1-(sinθ)^2]/(sinθ)^2 dθ
=∫(π/4, π/2)[1/(sinθ)^2]dθ-∫(π/4, π/2)dθ
=-cotθ(π/4, π/2)-θ(π/4, π/2)
=-cot(π/2)+cot(π/4)-π/2+π/4
=0+1-π/4
=1-π/4
∴当x=1时,θ=π/2, 当x=1/√2时,θ=π/4。
∴原式=∫(π/4, π/2) √[1-(sinθ)^2]/(sinθ)^2 cosθdθ
=∫(π/4, π/2)[(cosθ)^2/(sinθ)^2]dθ
=∫(π/4, π/2) [1-(sinθ)^2]/(sinθ)^2 dθ
=∫(π/4, π/2)[1/(sinθ)^2]dθ-∫(π/4, π/2)dθ
=-cotθ(π/4, π/2)-θ(π/4, π/2)
=-cot(π/2)+cot(π/4)-π/2+π/4
=0+1-π/4
=1-π/4
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询