(∣sinx∣+∣cosx∣)2=1+∣sin2x∣, which has period π/2... more directly, 1+∣sin2x∣ has period π, so with 20π=20 periods:
\[I = 20\int_0^\pi(1+|\sin2x|)\,dx = 40\int_0^{\pi/2}(1+\sin2x)\,dx = 40\left[x-\frac{\cos2x}2\right]_0^{\pi/2} = 40\left(\frac\pi2+1\right) = 20(\pi+2)\]