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解:
令∫(0,π) xsinx/(1+cos²x) dx = I,再令:x=π-t,则:0≤t≤π.
I=∫(0,π)(π-t)sin(π-t)/[1+cos²(π-t)]d(π-t)
=∫(π,0)(π-t)sint/(1+cos²t)dt
=π∫(0,π)dcost/(1+cos²t)-∫(π,0)tsint/(1+cos²t)dt
=π∫(0,π)dcost/(1+cos²t) - I
∴
I = (π/2)∫(0,π)dcost/(1+cos²t)
=(π/2)∫(0,π)dcosx/(1+cos²x)
令∫(0,π) xsinx/(1+cos²x) dx = I,再令:x=π-t,则:0≤t≤π.
I=∫(0,π)(π-t)sin(π-t)/[1+cos²(π-t)]d(π-t)
=∫(π,0)(π-t)sint/(1+cos²t)dt
=π∫(0,π)dcost/(1+cos²t)-∫(π,0)tsint/(1+cos²t)dt
=π∫(0,π)dcost/(1+cos²t) - I
∴
I = (π/2)∫(0,π)dcost/(1+cos²t)
=(π/2)∫(0,π)dcosx/(1+cos²x)
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