设f(x)连续,证明(积分区间为0到π)∫xf(sinx)dx=(π/2)∫f(sinx)dx
人气:359 ℃ 时间:2019-08-18 02:37:33
解答
证明:令x=π-t,则x由0到π,t由π到0,dx=-dt
原式记为I
则I=-(积分区间π到0)∫(π-t)f(sin(π-t)dt
=-(积分区间π到0)∫(π-t)f(sin(t)dt
=(积分区间0到π)∫(π-t)f(sin(t)dt
=(积分区间0到π)∫πf(sin(t)dt-I
所以2I=(积分区间0到π)∫πf(sin(t)dt
即I=(π/2)∫f(sint)dt=(π/2)∫f(sinx)dx
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