Question

求打勾两题答案谢谢了

Answer 1

2
√x=y
x=y^2
原式=∫(0,2)[(1-y)/(1+y)]d(y^2)
=-2∫(0,2)[y(y-1)/(y+1)]d(y)
=-2∫(0,2){(y-2)+[2/(y+1)]}d(y)
=-2{(y^2/2-2y)+2ln(y+1)}|(0,2)
=-2[(2-4+2ln3)-(0-0+0)]
=4-4ln3

追问

第二题谁对你看下？

追答

3
x=siny
原式=∫(0,π/2){√[1-(siny)^2]}d(siny)
=∫(0,π/2)[(cosy)^2]dy
=∫(0,π/2)[(cosy)^2]dy
=(1/2)*∫(0,π/2)(1+cos2y)dy
=(1/2)*(y-1/2*sin2y) | (0,π/2)
=(1/2)*[(π/2-0)-(0-0)]
=π/4

追问

你看下第二题不知道哪个对

谢谢了

Answer 2

(2)
let
u=√x
2udu= dx
x=0, u=0
x=4, u=2
∫(0->4) (1-√x)/(1+√x) dx
=∫(0->2) [(1-u)/(1+u) ](2udu)
=2∫(0->2) [(u-u^2)/(1+u) ] du
=2∫(0->2) [-u +2 - 2/(1+u) ] du
=2[-(1/2)u^2 +2u -2ln|1+u|] |∫(0->2)
= 2(-2 +4 -3ln3)
=4 -6ln3
(3)
x=sinu
dx=cosu du
x=0, u=0
x=1, u=π/2
-------
∫(0->1)√(1-x^2) dx
=∫(0->π/2) (cosu)^2 du
=(1/2)∫(0->π/2) (1+cos2u) du
=(1/2)[u+(1/2)sin2u]|(0->π/2)
=π/4

追问

你第二题对？

追答

(2)
let
u=√x
2udu= dx
x=0, u=0
x=4, u=2
∫(0->4) (1-√x)/(1+√x) dx
=∫(0->2) [(1-u)/(1+u) ](2udu)
=2∫(0->2) [(u-u^2)/(1+u) ] du
=2∫(0->2) [-u +1 - 1/(1+u) ] du
=2[-(1/2)u^2 +u -ln|1+u|] |∫(0->2)
= 2(-2 +2 -3ln3)
= -6ln3

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谢

Answer 3

3 - 2 Ln[4]
π/4

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要过程

谢谢了

Answer 4

令x=sint ,dx=costdt 当x=0时,t=0,当x=1时,t=π/2
∫（0,1）根号下[1-x^2]dx
= ∫(0,π/2)√(1-sin²t)costdt
= ∫(0,π/2)√(1-sin²t)costdt
=∫(0,π/2)cos²tdt
=∫(0,π/2) [(cos2t +1)/2]dt
=1/2∫(0,π/2)cos2t+1dt
=1/2[1/2∫(0,π/2)cos2td2t+∫(0,π/2)dt]
=1/2[1/2(sinπ-sin0)+(π/2-0)]
=1/2×π/2
=π/4 这样吗 3题

追问

第二题了？

Answer 5

3 - 2 Ln[4]