∫(0,π)|sinx-cosx|dx

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∫(0,π)|sinx-cosx|dx

∫(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)](π/4,π)
=√2-1-[(-1)-√2]
=2√2

∫π0(sinx+cosx)dx ∫(0,π)|sinx-cosx|dx =∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx ∫cosx / (cosx+sinx)dx ∫dx/（sinx+cosx) ∫(sinx-cosx)dx ∫sinx/(sinx-cosx)dx ∫(0,π/2)(-sinx+cosx)/(sinx+cosx)dx 请用换元法求出定积分 ∫【0到π/2】(sinx^10-cosx^10)dx/(5-sinx-cosx) ∫(0,π)x(e^sinx)|cosx|dx 定积分∫（0~π）（sinx+cosx）dx, ∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2]