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(1)
let
u=π/2-x
du=-dx
x=0, u=π/2
x=π/2, u=0
∫(0->π/2) f(sinx) dx
=∫(π/2->0) f(cosu) (-du)
=∫(0->π/2) f(cosu) du
=∫(0->π/2) f(cosx) dx
(2)
let
u=π-x
du = -dx
x=0, u=π
x=π, u=0
∫(0->π) xf(sinx) dx
=∫(π->0) (π-u)f(sinu) -du
=∫(0->π) (π-x)f(sinx) dx
2∫(0->π) xf(sinx) dx =∫(0->π) πf(sinx) dx
∫(0->π) xf(sinx) dx =(π/2)∫(0->π) f(sinx) dx
f(sinx ) = sinx/[1+(cosx)^2]
∫(0->π) xsinx/[1+(cosx)^2] dx
=∫(0->π) πf(sinx) dx
=(π/2)∫(0->π) f(sinx) dx
=(π/2)∫(0->π) sinx/[1+(cosx)^2] dx
=0
let
y =π-x
x=0, y=π
x=π, y=0
dy =-dx
∫(0->π) sinx/[1+(cosx)^2] dx
=∫(π->0) siny/[1+(cosy)^2] (-dy)
=-∫(0->π) sinx/[1+(cosx)^2] dx
2∫(0->π) sinx/[1+(cosx)^2] dx =0
∫(0->π) sinx/[1+(cosx)^2] dx =0
let
u=π/2-x
du=-dx
x=0, u=π/2
x=π/2, u=0
∫(0->π/2) f(sinx) dx
=∫(π/2->0) f(cosu) (-du)
=∫(0->π/2) f(cosu) du
=∫(0->π/2) f(cosx) dx
(2)
let
u=π-x
du = -dx
x=0, u=π
x=π, u=0
∫(0->π) xf(sinx) dx
=∫(π->0) (π-u)f(sinu) -du
=∫(0->π) (π-x)f(sinx) dx
2∫(0->π) xf(sinx) dx =∫(0->π) πf(sinx) dx
∫(0->π) xf(sinx) dx =(π/2)∫(0->π) f(sinx) dx
f(sinx ) = sinx/[1+(cosx)^2]
∫(0->π) xsinx/[1+(cosx)^2] dx
=∫(0->π) πf(sinx) dx
=(π/2)∫(0->π) f(sinx) dx
=(π/2)∫(0->π) sinx/[1+(cosx)^2] dx
=0
let
y =π-x
x=0, y=π
x=π, y=0
dy =-dx
∫(0->π) sinx/[1+(cosx)^2] dx
=∫(π->0) siny/[1+(cosy)^2] (-dy)
=-∫(0->π) sinx/[1+(cosx)^2] dx
2∫(0->π) sinx/[1+(cosx)^2] dx =0
∫(0->π) sinx/[1+(cosx)^2] dx =0
追问
你好可以看下第二题么
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