这个如何求积分?
1个回答
展开全部
∫√郑渗[(1-p^2)/(1+p^2)]pdp
=(1/2)∫√[(1-p^2)/(1+p^2)]dp^2
p^2=cosu
=(1/2)∫tan(u/2)dcosu
=(-1/2)∫sin(u/2)^2du
=(1/4)∫(cosu-1)du
=(1/4)(sinu-u)+C
=(1/4)(√(1-p^2)-arccos(p^2))+C
∫[0,1]√[(1-p^2)/(1+p^2)]pdp
=(1/4)(0-arccos1)-(1/4)(1-arccos0)
=(-1/4)(1-π/或丛晌衫锋2)
=π/8-1/4
=(1/2)∫√[(1-p^2)/(1+p^2)]dp^2
p^2=cosu
=(1/2)∫tan(u/2)dcosu
=(-1/2)∫sin(u/2)^2du
=(1/4)∫(cosu-1)du
=(1/4)(sinu-u)+C
=(1/4)(√(1-p^2)-arccos(p^2))+C
∫[0,1]√[(1-p^2)/(1+p^2)]pdp
=(1/4)(0-arccos1)-(1/4)(1-arccos0)
=(-1/4)(1-π/或丛晌衫锋2)
=π/8-1/4
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