# integration from 0 to pi |1/2 + cos x | dx can someone please help me with this

Examine the graph of 1/2 + cosx from x=0 to $\pi$, you will find that it is negative for x $\small \epsilon$ (2$\small \pi$/3 , $\small \pi$]
$\dpi{100} \small \therefore$ $\dpi{100} \small \int_{0}^{\pi}$|1/2 + cosx| = $\dpi{100} \small \int_{0}^{2\pi /3}$(1/2 + cosx)  –  $\dpi{100} \small \int_{2\pi /3}^{\pi}$  (1/2 + cosx)
= $\dpi{100} \small \[x/2 + sinx]_{0}^{2\pi /3}$  –  $\dpi{100} \small \[x/2 + sinx]_{2\pi /3}^{\pi}$
= [($\dpi{100} \small \pi$/3 + sin2$\dpi{100} \small \pi$/3) – (0 + 0)] – [($\dpi{100} \small \pi$/2 + 0) – ($\dpi{100} \small \pi$/3 + sin2$\dpi{100} \small \pi$/3)]
= ($\dpi{100} \small \pi$/3 + $\dpi{80} \small \sqrt{3}$ / 2) – $\dpi{100} \small \pi$/2 + ($\dpi{100} \small \pi$/3 + $\dpi{80} \small \sqrt{3}$ / 2)   = 2$\dpi{100} \small \pi$/3 – $\dpi{100} \small \pi$/2 + $\dpi{80} \small \sqrt{3}$
=  $\dpi{100} \small \pi$/6 + $\dpi{80} \small \sqrt{3}$