Let I=∫04π1+∣cosx∣xdx. Using ∫0af(x)dx=∫0af(a−x)dx: I=∫04π1+∣cosx∣4π−xdx. Adding: 2I=4π∫04π1+∣cosx∣dx. Since ∣cosx∣ has period π, ∫04π=4∫0π=4⋅2=8. So 2I=32π, giving I=16π.
∫04π1+∣cosx∣xdx=
Held on 10 Aug 2022 · Verified 13 Jul 2026.
4π
16π
32π
64π
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