Question

高数题求答案

第5题，要过程

Answer 1

let

f(x)=x.√(1-x^2)^3

f(-x) =-f(x)

=>∫(-1->1) x.√(1-x^2)^3 dx =0

let

x=sinu

dx=cosu du

x=0, u=0

x=1, u=π/2

∫(-1->1) (8-x)√(1-x^2)^3 dx

=∫(-1->1) 8.√(1-x^2)^3 dx

=16∫(0->1) √(1-x^2)^3 dx

=16∫(0->π/2)  (cosu)^4 du

=4∫(0->π/2)  (1+cos2u)^2 du

=4∫(0->π/2)  [1+2cos2u+(cos2u)^2] du

=2∫(0->π/2)  [3+4cos2u+cos4u ] du

=2[3u+2sin2u+(1/4)sin4u]|(0->π/2)

=3π

Answer 2

5. I = 8∫<-1, 1>(1-x^2)^(3/2)dx - ∫<-1, 1>x(1-x^2)^(3/2)dx
= 16∫<0, 1>(1-x^2)^(3/2)dx - 0 , 令 x = sint
= 16∫<0, π/2>(cost)^4dt = 4∫<0, π/2>(1+cos2t)^2dt
= 4∫<0, π/2>[1+2cos2t+(cos2t)^2]dt
= 2∫<0, π/2>[3+4cos2t+cos4t]dt
= 2[3t + 2sin2t + (1/4)sin4t]<0, π/2>
= 3π