Question

高等数学定积分？

π/2哪来的？

Answer 1

let
u=π-x
x=0, u=π
x=π, u=0
∫(0->π) x(sinx)^4 dx
=∫(π->0) (π-u)(sinu)^4 (-du)
=∫(0->π) (π-u)(sinu)^4 du
=∫(0->π) (π-x)(sinx)^4 dx
2∫(0->π) x(sinx)^4 dx =π∫(0->π) (sinx)^4 dx
∫(0->π) x(sinx)^4 dx =(π/2)∫(0->π) (sinx)^4 dx
let
I(2n)
=∫(0->π) (sinx)^(2n) dx
=-∫(0->π) (sinx)^(2n-1) dcosx
=-[cosx.(sinx)^(2n-1)]|(0->π) +(2n-1)∫(0->π) (sinx)^(2n-2) (cosx)^2 dx
=0+(2n-1)∫(0->π) (sinx)^(2n-2) [1-(sinx)^2] dx
2nI(2n) = (2n-1)I(2n-2)
I(2n) = [(2n-1)/(2n)] I(2n-2)
(1/π^2)∫(0->π) x(sinx)^4 dx

=(1/π^2) (π/2)∫(0->π) (sinx)^4 dx
=[1/(2π)]∫(0->π) (sinx)^4 dx
=[1/(2π)]I4
=[1/(2π)] (3/4)I2
=[1/(2π)] (3/4)(1/2)I0
=[1/(2π)] (3/4)(1/2)(π)
=3/16