用公式∫(0.π)xf(sinx)dx=π/2∫(0.π)f(sinx)dx计算：∫（0,π）(xsinx)/[1+(c

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用公式∫(0.π)xf(sinx)dx=π/2∫(0.π)f(sinx)dx计算：∫（0,π）(xsinx)/[1+(cosx)^2]dx
(0,π)中,0是下限,π是上限,答案是(π^2)/4,求详解

∫[0,π] (x sinx)/(1 + cos²x) dx
= ∫[0,π] (x sinx)/(2 - sin²x) dx,设f(x) = x/(2 - x²),则f(sinx) = sinx/(2 - sin²x)
= ∫[0,π] x f(sinx) dx
= (π/2)∫[0,π] f(sinx) dx
= (π/2)∫[0,π] sinx/(2 - sin²x) dx
= -(π/2)∫[0,π] 1/(1 + cos²x) d(cosx)
= -(π/2)arctan(cosx)_[0,π]
= -(π/2)[arctan(-1) - arctan(1)]
= -(π/2)(-π/4 - π/4)
= π²/4
之前应该还有个问题,证明这类型的积分适用于这条公式的.

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