Question

求解这个积分！！！

如图

Answer 1

∫<√(x^2+2) -> √(4-x^2) > dy

=√(4-x^2) -√(x^2+2)

∫(-1->1) [∫<√(x^2+2) -> √(4-x^2) > dy ] dx

=∫(-1->1) [√(4-x^2) -√(x^2+2) ] dx

=2∫(0->1)√(4-x^2)  dx-  2∫(0->1) √(x^2+2)  dx

=2π -2[√3/2 + ln|(√6+√2)/2| ]

=2π -√3 - ln|(√6+√2)/2| ]

let

x=2sinu

dx=2cosu du

x=0, u=0

x=1, u=π/2

∫(0->1)√(4-x^2)  dx

=4∫(0->π/2) (cosu)^2 du

=2∫(0->π/2) (1+cos2u) du

=2[ u +(1/2)sin2u]|(0->π/2)

=π

/

let

x= √2tanv

dx=√2(secv)^2 dx

x=0, v=0

x=1, v=arctan(1/√2)

∫(0->1) √(x^2+2)  dx

=2∫(0->arctan(1/√2))  (secv)^3 dv

=[secv.tanv +ln|secv+tanv| ] |(0->arctan(1/√2)) 

=√3/2 + ln|(√6+√2)/2| 

/

∫(secv)^3 dv

=∫secv dtanv

=secv.tanv -∫(tanv)^2.secv dv

=secv.tanv -∫[(secv)^2-1].secv dv

2∫(secv)^3 dv =secv.tanv +∫secv dv

∫(secv)^3 dv =(1/2)[secv.tanv +ln|secv+tanv| ] +C

追问

u不是应该等于π/6？

追答

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