∫[0,π]√(1+cos2x)dx=∫[0,π]√(1+2cos²-1)dx=√2∫[0,π]|cosx|dx=√2∫[0,π/2]cosxdx+√2∫[π/2,π](-sinx)dx=√2(sinx [0,π/2])-√2(sinx[π/2,π])=√2(1-0)-√2(0-1)=2√2