Question

定积分有换元法和分部积分法 凑微分又是什么呢

Answer 1

(π/2)∫(0->π) [ xsinx /(1+(cosx)^2 ) ]dx
let
y = π-x
dy = -dx
x=0, y=π
x=π,y=0
∫(0->π) [ xsinx /(1+(cosx)^2 ) ]dx =∫(π->0) [ ∫(π->0) [ (π-y)siny /(1+(cosy)^2 ) ](-dy)
=∫(0->π) [ ∫(0->π) [ (π-x)sinx /(1+(cosx)^2 ) ]dx
2∫(0->π) [ xsinx /(1+(cosx)^2 ) ]dx =π∫(0->π) sinx/(1+(cosx)^2 ) ]dx
∫(0->π) [ xsinx /(1+(cosx)^2 ) ]dx = (π/2)∫(0->π) sinx/(1+(cosx)^2 ) ]dx
(π/2)∫(0->π) [ xsinx /(1+(cosx)^2 ) ]dx =(π/2)^2∫(0->π) sinx/(1+(cosx)^2 ) ]dx

追问

我不懂 能用文字说明吗
谢谢