Question

有人会这道积分题吗？

比较懵，过程最好详细点儿

Answer 1

Answer 2

用变量替换方法积分即可，给出关键步骤，最后你可以计算答案了

Answer 3

乱七八糟答案真多……详细过程如图rt所示……希望能帮到你解决你心中的问题

Answer 4

如下图所示，一开始用分部积分，然后再用换元法做

Answer 5

let

x=(3/5)sinu
dx=(3/5)cosu du
x=0, u=0
x=0.6, u=π/2
∫(0->0.6) x^2/√(9-25x^2) dx
=∫(0->π/2) [(9/25)(sinu)^2/(3/5)cosu] . [(3/5)cosu du ]
=(9/25)∫(0->π/2) (sinu)^2 du
=(9/50)∫(0->π/2) (1-cos2u) du
=(9/50) [u-(1/2)sin2u]|(0->π/2)
=(9/100)π

Answer 6

∫<0,0.6>[x²/√(9-25x²)]dx=(1/3)∫<0,0.6>x²dx/√[1-(5x/3)²]
【 x=(3/5)sinu，dx=(3/5)cosudu；x=0时u=0；x=0.6时u=π/2】
=(1/3)∫<0,π/2>[(9/25)sin²u]/√(1-sin²u)](3/5)cosudu
=(9/125)∫<0,π/2>sin²udu=(9/250)∫<0,π/2>(1-cos2u)du
=(9/250)[u-(1/2)sin2u]<0,π/2>=9π/500;