Question
The value of $$\int_0^\pi {\ln \left( {1 + \cos \,x} \right)} dx$$ is :
A.
$$\frac{\pi }{2}\log \,2$$
B.
$$\pi \,\log \,2$$
C.
$$ - \pi \,\log \,2$$
D.
$$0$$
Answer :
$$ - \pi \,\log \,2$$
Solution :
$$\eqalign{
& I = \int_0^\pi {\log \left( {1 + \cos \,x} \right)} dx \cr
& = \int_0^\pi {\log \left( {2\,{{\cos }^2}\frac{x}{2}} \right)} dx \cr
& = \int_0^\pi {\left( {\log \,2 + 2\,\log \,\cos \frac{x}{2}} \right)} dx \cr
& = \int_0^\pi {\log \,2\,dx + 2\int_0^\pi {\log \,\cos \frac{x}{2}dx} } \cr
& = \pi \,\log \,2 + 2\int_0^{\frac{\pi }{2}} {\left( {2\,\log \,\cos \,t} \right)} dt\,\,\,\,\,\left( {{\text{where }}\frac{x}{2} = t} \right) \cr
& = \pi \,\log \,2 + 4\left( { - \frac{\pi }{2}\log \,2} \right) \cr
& = \pi \,\log \,2 - 2\pi \,\log \,2 \cr
& = - \pi \,\log \,2 \cr
& \left[ {\int_0^{\frac{\pi }{2}} {\log \,\sin \,\theta \,d\theta = } \int_0^{\frac{\pi }{2}} {\log \,\cos \,\theta \,d\theta } = - \frac{\pi }{2}\log \,2} \right] \cr} $$