用公式∫(0.π)xf(sinx)dx=π/2∫(0.π)f(sinx)dx计算:∫(0,π)(xsinx)/[1+(cosx)^2]dx
问题描述:
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
答
∫[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...