换元法算定积分，步骤，谢谢

Question

3，5，7题

Answer 1

如下

Answer 2

5.令x=sinz,dx=cosz dz,cosz=√(1-x²)
∫ x²/√(1-x²) dx = ∫ sin²z*cosz/√(1-sin²z) dz
= ∫ sin²z*cosz/cosz dz
= ∫ sin²z dz
= (1/2)∫ (1-cos2z) dz
= (1/2)(z-1/2*sin2z) + C
= (1/2)z-1/2*sinz*cosz + C
= (1/2)arcsinx - 1/2*x*√(1-x²) + C
= (1/2)[arcsinx - x√(1-x²)] + C追答

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Answer 3

(3)

let
x=tanu
dx = (secu)^2 du
x=0, u=0
x=1, u=π/4
∫(0->1) x^2/(1+x^2)^3 dx
=∫(0->π/4) [(tanu)^2/(secu)^6] [( secu)^2 du]
=∫(0->π/4) [(tanu)^2/(secu)^4] du
=∫(0->π/4) (sinu.cosu)^2 du
=(1/4)∫(0->π/4) (sin2u)^2 du
=(1/8)∫(0->π/4) (1-cos4u) du
=(1/8)[u -(1/4)sin4u]|(0->π/4)
=π/32
(5)
let
x= sinu
dx= cosu du
x=0, u=0
x=1, u=π/2
∫(0->1) x^2. √(1-x^2) dx
=∫(0->π/2) (sinu.cosu)^2 du
=(1/4)∫(0->π/2) (sin2u)^2 du

=(1/8)∫(0->π/2) (1-cos4u) du
=(1/8)[u -(1/4)sin4u]|(0->π/2)
=π/16
(7)
let
lnx = (tanu)^2
(1/x)dx = 2 tanu.(secu)^2 du
x=1, u=0
x=e^3 , u=arctan(3)
∫(1->e^3) dx/ [x.√(1+lnx) ]
=∫(0->arctan(3) ) 2 tanu.(secu)^2 du/(secu)
=2∫(0->arctan(3) ) tanu.secu du
=2[ secu]|(0->arctan(3) )
= 2( √10 - 1)