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高等数学,求大神帮忙看一下这道题,
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∫(nπ/2,(n+1)π/2)f(|sinx|)dx
当n为偶数时,设n=2k,则∫(kπ,(kπ+π/2))f(|sinx|)dx 设x=kπ+t
=∫(0,π/2)f(|sin(kπ+t)|)d(kπ+t)
=∫(0,π/2)f(|sint|)dt
当n为奇数时,设n=2k+1,则∫(nπ/2,(n+1)π/2)f(|sinx|)dx
=∫(kπ+π/2,(k+1)π)f(|sinx|)dx 设x=kπ+t+π
=∫(-π/2,0)f(|sin(kπ+t+π)|)d(kπ+t+π)
=∫(-π/2,0)f(|sint|)dt
=∫(0,π/2)f(|sint|)dt
即对于任意正整数n,均有:∫(nπ/2,(n+1)π/2)f(|sinx|)dx=∫(0,π/2)f(|sint|)dt
当n为偶数时,设n=2k,则∫(kπ,(kπ+π/2))f(|sinx|)dx 设x=kπ+t
=∫(0,π/2)f(|sin(kπ+t)|)d(kπ+t)
=∫(0,π/2)f(|sint|)dt
当n为奇数时,设n=2k+1,则∫(nπ/2,(n+1)π/2)f(|sinx|)dx
=∫(kπ+π/2,(k+1)π)f(|sinx|)dx 设x=kπ+t+π
=∫(-π/2,0)f(|sin(kπ+t+π)|)d(kπ+t+π)
=∫(-π/2,0)f(|sint|)dt
=∫(0,π/2)f(|sint|)dt
即对于任意正整数n,均有:∫(nπ/2,(n+1)π/2)f(|sinx|)dx=∫(0,π/2)f(|sint|)dt
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