2个回答
展开全部
∫dx/√[(a²-x²)³]
=∫dx/(a²-x²)^(3/2)
设x=asint,dx=acostdt
原式=∫acostdt/[a²-(asint)²]^(3/2)
=∫acostdt/[(acost)²]^(3/2)
=∫acostdt/(acost)³
=∫dt/(a²cos²t)
=1/a²*tant+C
=1/a²*sint/√(1-sin²t)+C
=1/a²*asint/√(a²-a²sin²t)+C
=1/a²*x/√(a²-x²)+C
=∫dx/(a²-x²)^(3/2)
设x=asint,dx=acostdt
原式=∫acostdt/[a²-(asint)²]^(3/2)
=∫acostdt/[(acost)²]^(3/2)
=∫acostdt/(acost)³
=∫dt/(a²cos²t)
=1/a²*tant+C
=1/a²*sint/√(1-sin²t)+C
=1/a²*asint/√(a²-a²sin²t)+C
=1/a²*x/√(a²-x²)+C
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