Question

高等数学 定积分计算题？

Answer 1

Answer 2

∫(tanx)^2secxdx = ∫[(secx)^2-1]secxdx = ∫(secx)^3dx - ∫secxdx
= ∫secxdtanx - ∫secxdx = secxtanx - ∫(tanx)^2secxdx - ln|secx+tanx|
则 2∫(tanx)^2secxdx = secxtanx - ln|secx+tanx|
I = 4π∫<0, √5>(tanx)^2secxdx = 2π[secxtanx - ln|secx+tanx|]<0, √5>
= 2π[sec√5tan√5 - ln|sec√5+tan√5|]

Answer 3

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