Evaluate the following integral I=∫_0^πln⁡(1+cos⁡θ ) dθ

I=∫_0^πln⁡(2〖cos〗^2  θ/2)  dθ =∫_0^π[ln⁡2+ln⁡(〖cos〗^2  θ/2) ]  dθ
=∫_0^πln⁡2  dθ+2∫_0^πlog⁡cos⁡〖θ/2〗   dθ=π ln⁡2+2∫_0^πln⁡cos⁡〖θ/2〗   dθ
=π ln⁡2+4∫_0^(π/2)ln⁡cos⁡∅   dθ  ,where   θ/2=∅
=π ln⁡2-4.π/2  ln⁡2=-π ln⁡2
where  ∫_0^(π/2)ln⁡cos⁡x   dx=-π/2  ln⁡2=π/2  ln⁡〖1/2〗