∫(0,π)|sinx-cosx|dx

来源：学生作业帮助网编辑：作业帮时间：2024/05/02 21:18:18

∫(0,π)|sinx-cosx|dx
∫(0,π)|sinx-cosx|dx

∫(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

∫(0,π)|sinx-cosx|dx ∫dx/（sinx+cosx) ∫(sinx-cosx)dx ∫cosx / (cosx+sinx)dx ∫sinx/(sinx-cosx)dx ∫sinx/(cosx-sinx )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) 定积分∫（0~π）（sinx+cosx）dx, ∫(0→π)(sinx+cosx)dx值多少 ∫(0~π) sinx/(5 - 4cosx) dx怎么求? ∫(sinx+cosx)^2 dx ∫(sinx+cosx)^2 dx ∫sinxcosx/(sinx+cosx)dx ∫(sinx)^3/(cosx)dx ∫/(1+sinx+cosx)dx ∫cos2x/(sinx+cosx)dx ∫（cosx/1+sinx)dx ∫(cosx/e^sinx)dx