Question

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

Answer 1

∫[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。

Answer 2

确的常