请教一位高数高手来解答这三道求定积分题:
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发布时间：2024-02-19 15:05
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热心网友
时间：2024-07-22 23:20
=∫(-2,0) 1/[(x+1)²+1] d(x+1)
=arctan(x+1)|(-2,0)
=arctan1 - arctan(-1)
=π/2
2.∫(0,π) (1-sin³x)dx
=∫(0,π) dx - ∫(0,π) sin³xdx
=π + ∫(0,π) sin²xdcosx
=π + ∫(0,π) (1-cos²x)dcosx
=π + (cosx - cos³x/3)|(0,π)
=π - 4/3
3.∫(0,π) √(sinx-sin³x)dx
=∫(0,π) cosx√sinxdx
=∫(0,π) √sinxdsinx
=(2sinx√sinx)/3 |(0,π)
=0
