Question

这题定积分怎么求？

Answer 1

Answer 2

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