求解一道定积分问题 20
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∫(0,π)√[sinx-(sinx)^3]dx
=∫(0,π)√[sinx(cosx)^2]
=∫(0,π/2)cosx√sinxdx-∫(π/2,π)cosx√sinxdx
=∫(0,π/2)√sinxdsinx-∫(π/2,π)√sinxdsinx
=(2/3)(sinx)^(3/2)-(2/3)(sinx)^(3/2)
=(2/3)(sinπ/2)^(3/2)-(2/3)(sin0)^(3/2)-(2/3)(sinπ)^(3/2)+(2/3)(sinπ/2)^(3/2)
=2/3-0-0+2/3
=4/3
=∫(0,π)√[sinx(cosx)^2]
=∫(0,π/2)cosx√sinxdx-∫(π/2,π)cosx√sinxdx
=∫(0,π/2)√sinxdsinx-∫(π/2,π)√sinxdsinx
=(2/3)(sinx)^(3/2)-(2/3)(sinx)^(3/2)
=(2/3)(sinπ/2)^(3/2)-(2/3)(sin0)^(3/2)-(2/3)(sinπ)^(3/2)+(2/3)(sinπ/2)^(3/2)
=2/3-0-0+2/3
=4/3
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你这不是我说的那道题啊
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