请教一道证明题
记住:定积分的结果只与被积函数、积分上下限有关,与被积变量的形式无关
设u=π -x代入原式
∫xf(sinx)dx
=∫(π -u)f(sinu)du[积分区间0->π]
=∫(π -x)f(sinx)dx[积分区间0->π]
=π∫f(sinx)dx-∫xf(sinx)dx[积分区间0->π]
=〉2∫xf(sinx)dx =π∫f(sinx)dx[积分区间0->π]
=〉∫xf(sinx)dx=(π/2)*∫f(sinx)dx[积分区间0->π]
在第一道题的基础之上
∫xf(sinx)dx=(π/2)*∫f(sinx)dx[积分区间0->π/2]
+(π/2)*∫f(sinx)dx...全部
记住:定积分的结果只与被积函数、积分上下限有关,与被积变量的形式无关
设u=π -x代入原式
∫xf(sinx)dx
=∫(π -u)f(sinu)du[积分区间0->π]
=∫(π -x)f(sinx)dx[积分区间0->π]
=π∫f(sinx)dx-∫xf(sinx)dx[积分区间0->π]
=〉2∫xf(sinx)dx =π∫f(sinx)dx[积分区间0->π]
=〉∫xf(sinx)dx=(π/2)*∫f(sinx)dx[积分区间0->π]
在第一道题的基础之上
∫xf(sinx)dx=(π/2)*∫f(sinx)dx[积分区间0->π/2]
+(π/2)*∫f(sinx)dx[积分区间π/2->π]
设u=π-x,∫f(sinx)dx[积分区间π/2->π]
=∫f(sinx)dx[积分区间0->π/2]
=>∫xf(sinx)dx=π*∫f(sinx)dx 积分区间为(0-π/2)
设u= π/2 - x代入原式
∫xf(sinx)dx
=∫(π/2 -u)f(cosu)du[积分区间-π/2->π/2]
=π/2∫f(cosx)dx-∫xf(cosx)dx[积分区间-π/2->π/2]
其中∫xf(cosx)dx=0[积分区间-π/2->π/2](关于x轴对称[xf(cosx)为奇函数])
π/2∫f(cosx)dx=π∫f(cosx)dx[积分区间0->π/2]](关于x轴对称[f(cosx)为偶函数])
=>∫xf(sinx)dx =π∫f(cosx)dx[积分区间0->π/2]
。
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