Question

帮忙算一下这个积分 ，不是很会，谢谢了

Answer 1

。侍迅兄老袭昌滑

追问

这儿不是很懂，虽然是对称积分，但是自变量是r啊

Answer 2

( 1+r^2.cosθsinθ)r

= r+r^3.cosθsinθ

=rcosθsinθ .(1+r^2)  +r(1-cosθsinθ)

∫(0->1) [ ( 1+r^2.cosθsinθ)/(1+r^2)] r dr

=∫(0->1) rcosθsinθ dr +  ∫(0->1)  r(1-cosθsinθ)/(1+r^2) dr

=(1/2)cosθsinθ[ r^2]|(0->1) +(1/2)(1-cosθsinθ)[ln|1+r^2|]|(0->1)

=(1/2)cosθsinθ + (1/2)(1-cosθsinθ)ln2

=(1/4)sin2θ + (1/2)ln2 - (1/4)ln2. sin2θ 

=(1/4)(1-ln2)sin2θ + (1/2)ln2

∫(-π/2->π/2) [∫晌和(0->没谨升1) [ ( 1+r^2.cosθsinθ)/(1+r^2)] r dr] dθ

=∫(-π/2->π/2) [ (1/4)(1-ln2)sin2θ + (1/2)ln2 ] dθ

=(1/2)ln2∫(-π/2->π/2) 枯老  dθ

=(1/2)ln2 [θ] |(-π/2->π/2)

=(1/2)(ln2)π