∫(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
