∫0π log⁡(1+cos⁡x)dx is equal to

Let,I=∫0π log⁡(1+cos⁡x)dx----iI=∫0π log⁡{1+cos⁡(π−x)}dx=∫0π log⁡(1−cos⁡x)dx[∵cos⁡(π−x)=−cos⁡x]… (ii) =∫0π log⁡2sin2⁡x2dx ∵1−cos⁡x=2sin2⁡x2=∫0π log⁡2+2log⁡sin⁡x2dx=∫0π log⁡2dx+2∫0π log⁡sin⁡x2dxIn the second. integral, put  x2=t⇒dx=2dtand limits when x =0, t =0 and when x=π,t=π/2∴ I=log⁡2[x]0π+2∫0π/2 log⁡(sin⁡t)2dt=(log⁡2)(π−0)+4−π2log⁡2∵∫0π/2 log⁡sin⁡xdx=−π2log⁡2=−πlog⁡2