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求问一道定积分
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∫(0,π)√(1-sinx)dx
=∫(0,π)√(1-2sinx/2cosx/2)dx
=∫(0,π)√(sinx/2-cosx/2)²dx
=∫(0,π)|sinx/2-cosx/2|dx
=2·2∫(0,π)(cosx/2-sinx/2)d(x/2)
=4(sinx/2+cosx/2)|(0,π)
=4(√2-1)
=∫(0,π)√(1-2sinx/2cosx/2)dx
=∫(0,π)√(sinx/2-cosx/2)²dx
=∫(0,π)|sinx/2-cosx/2|dx
=2·2∫(0,π)(cosx/2-sinx/2)d(x/2)
=4(sinx/2+cosx/2)|(0,π)
=4(√2-1)
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