Question

求解这个积分

如图 需要详细过程和步骤

Answer 1

∫√(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-2[√3/2+ln、(√6+√2)/2、]
=2π-√3-ln、(√6+√2)/2、]
let
x=2sinu
dx=2cosu
x=0，u=0
x=1，u=π/2
∫(0->1)√(4-x^2)dx
=4∫(0->π/2)(cosu)^2
=2∫(0->π/2)(1+cos2u)
=2[u+(1/2)sin2u]、(0->π/2)
=π
/
let
x=√2tanv
dx=√2(secv)^2dx
x=0，v=0
x=1，v=arctan(1/√2)
∫(0->1)√(x^2+2)dx
=2∫(0->arctan(1/√2))(secv)^3dv
=[secv.tanv+ln、secv+tanv、]、(0->arctan(1/√2))
=√3/2+ln、(√6+√2)/2、
/
∫(secv)^3dv
=∫secvdtanv
=secv.tanv-∫(tanv)^2.secvdv
=secv.tanv-∫[(secv)^2-1].secvdv
2∫(secv)^3dv=secv.tanv+∫secvdv
∫(secv)^3dv=(1/2)[secv.tanv+ln、secv+tanv、]+C。