$\int_{0}^{2\pi}{ln(1+\cos{x})}dx=$

$I = \int_0^{2\pi } {\ln \left( {1 + \cos x} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos x} \right)dx}$ ---(1) (${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$ if ${f\left( {2a - x} \right) = f\left( x \right)}$)

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos \left( {\pi - x} \right)} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 - \cos x} \right)dx}$ ---(2)

Adding (1) and (2) we get

$\Rightarrow 2I = 2\int_{}^{} {\ln \left[ {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} \right]dx}$

$\Rightarrow I = \int_0^\pi {\ln \left( {1 - {{\cos }^2}x} \right)dx}$

$\Rightarrow I = \int_0^\pi {\ln {{\sin }^2}xdx}$

$\Rightarrow I = 2\int_0^\pi {\ln \sin xdx}$ ---(A)

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin xdx}$ ---(3) (${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$${f\left( {2a - x} \right) = f\left( x \right)}$)

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin \left( {\frac{\pi }{2} - x} \right)dx}$

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {ln(cosx)dx}$ ---(4)

Adding (3) and (4)

$\Rightarrow 2I = 4\int_0^{\frac{\pi }{2}} {\ln \left( {\frac{{2\sin x\cos x}}{2}} \right)dx}$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\int_0^{\frac{\pi }{2}} {\ln 2dx} }$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\ln 2\left[ x \right]_0^{\frac{\pi }{2}}}$

putting $2x=t$ $\Rightarrow 2dx = dt$

$\Rightarrow I = \int_0^\pi {\ln \sin tdt} - 2\ln 2\left[ {\frac{\pi }{2}} \right]$

$\Rightarrow I = \int_0^\pi {\ln \sin x dx} - \pi \ln 2$

(since ${\int_a^b {f\left( t \right)dt = \int_a^b {f\left( x \right)dx} } }$)

$\Rightarrow I = \frac{I}{2} - \pi \ln 2$ (from (A))

$\Rightarrow \frac{I}{2} = - \pi \ln 2$

$\Rightarrow I = - 2\pi \ln 2$