证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx(0,π)中,0是下限,π是上限,后面同理,求详解

如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx 怎么证明∫(0到pi)f(sinx)dx=2*∫(0到pi/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 定积分证明题 ——请证明：【积分区间为0到π】∫xf(sinx)dx=(π/2)∫f(sinx)dx 证明:定积分∫（0到π）f(sinx)dx=2∫（0到π／2）f(sinx)dx, 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 设f（x）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)f（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)为连续函数,证明:∫(0,π)f(丨cosx丨)dx=2∫(0,π/2)f(sinx)dx 设f(x)连续,证明（积分区间为0到2π）∫xf(cosx)dx=π∫f(sinx)dx 求证：∫(0至π) x f(sinx)dx = π/2∫(0至π)f(sinx)dx f为连续函数 证明f(cosx)dx=f(sinx)dx 左右边的范围都是0到π /2 证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dxf(x)在区间[0,1]连续 ∫a^bfxdx表示a到b的定积分 谁能证明一下:∫0^π/2f(sinx)dx=∫0^π/2f(cosx)dx ∫0^π/2 sinx^ndx=∫0^π 证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx(0,π)中,0是下限,π是上限,后面同理,求详解 设f(u)在[-1,1]上连续,利用变换x=π-t,证明∫（π 0）xf(sinx)dx=π/2*∫（π 0）f(sinx)dx 积分∫f(sinx)/[f(cosx)+f(sinx)]dx= 在0到π/2的范围内