Question

定积分计算问题？

求解

Answer 1

(1)
let
u=π-x
du= -dx
x=0, u=π
x=π, u=0
∫(0->π) xsinx/[1+(sinx)^2] dx
=∫(π->0) { (π-u)sinu/[1+(sinu)^2] } (-du)
=∫(0->π) (π-u)sinu/[1+(sinu)^2] du
=∫(0->π) (π-x)sinx/[1+(sinx)^2] dx
2∫(0->π) xsinx/[1+(sinx)^2] dx =π∫(0->π) sinx/[1+(sinx)^2] dx
∫(0->π) xsinx/[1+(sinx)^2] dx
=(π/2)∫(0->π) sinx/[1+(sinx)^2] dx
=-(π/2)∫(0->π) dcosx/[1+(sinx)^2]
=-(π/2)∫(0->π) dcosx/[2-(cosx)^2]
=-(√2π/8)∫(0->π) { 1/[√2-cosx] + 1/[√2+(cosx)] } dcosx
=(√2π/8) [ ln|[√2-(cosx)]/[√2+cosx]|] |(0->π)
=(√2π/8) { ln[(√2+1)/(√2-1)] - ln[(√2-1)/(√2+1)] }
=(√2π/4) ln[(√2+1)/(√2-1)]
=(√2π/4) ln[(√2+1)^2]
=(√2π/2) ln(√2+1)

(2)
let
u=√x
2u du = dx
x=0, u=0
x=1, u=1
∫(0->1) arctan√x dx

=[xarctan√x]|(0->1) -(1/2) ∫(0->1) √x/(1+x) dx
=π/4 - ∫(0->1) u^2/(1+u^2) du
=π/4 - ∫(0->1) [ 1 - 1/(1+u^2) ] du
=π/4 - [ u -arctanu]|(0->1)
=π/4 - [ ( 1-π/4) ]
=π/2 -1