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数学
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计算:∫(0,π/2)√(1-sin2x)dx
人气:249 ℃ 时间:2020-02-02 21:43:12
解答
1-sin2x=(sinx-cosx)^2
∫(0,π/2)√(1-sin2x)dx
=∫(0,π/4)√(1-sin2x)dx+∫(π/4,π/2)√(1-sin2x)dx
=∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx
=(√2-1)+(√2-1)
=2√2-2
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