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被积函数连续,证明:∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
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被积函数连续,证明:∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
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答案和解析
换元法.
∫(0,π)f(sinx)dx
= ∫(0,π/2)f(sinx)dx+∫(π/2,π)f(sinx)dx
换元,将后式中的x换成π-t
= ∫(0,π/2)f(sinx)dx-∫(0,π/2,)f(sin(π-t))d(π-t)
=∫(0,π/2)f(sinx)dx+∫(0,π/2)f(sin(t))dt
=∫(0,π/2)f(sinx)dx+∫(0,π/2)f(sinx)dx
=2∫(0,π/2)f(sinx)dx
∴ ∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
∫(0,π)f(sinx)dx
= ∫(0,π/2)f(sinx)dx+∫(π/2,π)f(sinx)dx
换元,将后式中的x换成π-t
= ∫(0,π/2)f(sinx)dx-∫(0,π/2,)f(sin(π-t))d(π-t)
=∫(0,π/2)f(sinx)dx+∫(0,π/2)f(sin(t))dt
=∫(0,π/2)f(sinx)dx+∫(0,π/2)f(sinx)dx
=2∫(0,π/2)f(sinx)dx
∴ ∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
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