求证：∫(0至π) x f(sinx)dx = π/2∫(0至π)f(sinx)dx

证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx 设f(x)∈C[0,1],证明∫（π,0）*x*f(sinx)dx =π/2*∫（π,0）*f(sinx)dx ∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2] 证明：若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx (上限 π,下限 0) 设f(x)连续,证明（积分区间为0到π）∫xf(sinx)dx=(π/2)∫f(sinx)dx 如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx 证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dx 积分∫f(sinx)/[f(cosx)+f(sinx)]dx= 在0到π/2的范围内 ∫0~2π x|sinx|dx 若f(x)在[0,1]上连续,证明 ∫【上π/2下0】f(sinx)dx= ∫【上π/2下0】f(cosx)dx 设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（sinx）dx 设f(x)为连续函数,证明:∫(0,π)f(丨cosx丨)dx=2∫(0,π/2)f(sinx)dx