证明的话,令t=π-x即可,设左边=t
t=∫(0->π)xf(sinx)dx=∫(0->π)(π-t)f(sint)dt=∫(0->π)(π-x)f(sinx)dx=π∫(0->π)f(sinx)dx-∫(0->π)xf(sinx)dx
=π∫(0->π)f(sinx)dx-t
所以
t=∫(0->π)xf(sinx)dx=(π/2)∫(0->π)f(sinx)dx
然后令t=π-x, 可证明∫(π/2->π)f(sinx)dx=∫(0->π/2)f(sinx)dx
所以
∫(0->π)xf(sinx)dx=(π/2)∫(0->π)f(sinx)dx=(π/2)[2∫(0->π/2)f(sinx)dx]=π∫(0->π/2)f(sinx)dx
有这个结论,
原积分=π∫(0->π/2)[sinx/(1+(cosx)^2)]dx=-π∫(0->π/2)dcosx/[(1+(cosx)^2)]
=-πarctan(cosx) |(0->π/2)
=π^2/4