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∫(0,π)|sinx-cosx|dx
人气:346 ℃ 时间:2020-03-28 07:32:23
解答
∫(0,π)|sinx-cosx|dx=∫(0,π)√2|sin(x-π/4)|dx=-∫(0,π/4)√2sin(x-π/4)dx+∫(π/4,π)√2sin(x-π/4)dx=-[-√2cos(x-π/4)](0,π/4)+[-√2cos(x-π/4)](π/4,π)=[√2cos(x-π/4)](0,π/4)-[√2cos(x-π/4)]...
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