Question

高数求定积分？

求过程

Answer 1

如果把分式拆分成两项,第一项是偶函数,第二项是奇函数,那麼第二项的积分为0,第一项只需要在[0,1]上积分再乘以2就行.
令x=sint,t∈[0,π/2],则√(1-x²)=cost,dx=costdt
原式=2∫[0,π/2]2sin²t/(1+cost)*costdt
=4∫[0,π/2](1-cos²t)/(1+cost)*costdt
=4∫[0,π/2](1-cost)*costdt
=4∫[0,π/2]costdt-4∫[0,π/2]cos²tdt
=4-4*(x/2+1/4*sin2x)|[0,π/2]
=4-π

Answer 2

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