\(\int_{0}^{4π}\)\(\frac{x}{1+|cosx|}\)dx

16π

\(I=\int_{0}^{4π}\frac{x}{1+||cos(4π-x)||}dx\) .......(i)

\(I=\int_{0}^{4π}\frac{4π-x}{1+||cos(4π-x)||}dx\) .......(ii)

 Adding eq. (i) and (ii),

\(2I=\int_{0}^{4π}\frac{4π}{1+||cosx||}dx⇒(\int_{0}^{4π}\frac{1}{1+||cosx||}dx)2π\)

$8π\int_{0}^{π}\frac{1}{1+||cosx||}dx⇒8π(\int_{0}^{\frac{π}{2}}\frac{dx}{1+cosx}+\int_{\frac{π}{2}}^{π}\frac{dx}{1-cosx})$

$8π(tan(\frac{x}{2})|_{0}^{\frac{π}{2}}+(-cot\frac{x}{2})|_{\frac{π}{2}}^{π})$

$=8π×2=16π$