Question

integrate log(1+cosx) from 0 to pi

Answer 1

I =log(1+cosx ) in limit [0,π] -------------(1)

use f(x ) =f(π+0 -x )

I =log(1+cos(π-x))
=log(1-cosx ) ------------(2)

add equation (1) and (2)
2I =log(1+cosx)(1-cosx)
=log(1-cos^2x) =log(sin^2x)
=2log(sinx)

I =log(sinx) in limit [0,π]
if limit [0,π/2 ]
then ,
I =2log(sinx )
now,
I =2log(sinx) in limit [0,π/2]
again
f(x ) =f(π/2 -x )
so,
I =2log(cosx)

add both
2I =2log(sinx.cosx)

I =log(sin2x/2) in limit [0,π/2]

=log(sin2x) in [0,π/2] -[ln2] in [0,π/2]

I" =log(sin2x) in limit [0,π/2]
now ,
if limit [0,π]
then
I" =1/2 log(sinx) =1/2I

now ,
I =I/2 -π/2ln2
I = -πln2 (answer)

Answer 2

Let  x/2 = t
     x varies from 0 to π.   t varies from 0 to π/2.

f(x) dx = Ln (1+cos x) = Ln [2 cos² x/2 ]  dx
           = Ln 2 dx + 2*2* Ln (Cos t) dt

$I=\int \limits_{0}^{\pi} {f(x)} \, dx= \int \limits_{x=0}^{\pi} {Ln2 } \, dx+4*\int \limits_{t=0}^{\pi/2} {Ln(Cos\ t) } \, dt\\\\I=\pi\ Ln2+I1\\\\I1=\int \limits_{t=0}^{\pi/2} {Ln(Cos\ t) } \, dt=\int \limits_{t=0}^{\pi/2} {Ln(Sin\ t) } \, dt,..check\ t\ -\ \textgreater \ \pi/2-t\\\\I1+I1=\int \limits_{t=0}^{\pi/2} {Ln[(Cos\ t)*(sin\ t)] } \, dt\\\\2I1=\int \limits_{0}^{\pi/2} {Ln(sin2t)} \, dt-\int \limits_{0}^{\pi/2} {Ln2} \, dt\\\\2I1=\frac{1}{2}\int \limits_{w=0}^{\pi} {Ln(sinw)} \, dw-\frac{\pi}{2}Ln2,,...w=2t$

$So\ 2I1=-\frac{1}{2}*(2\ I1)-\frac{\pi}{2}Ln2\\\\I1=\frac{\pi}{2}Ln2\\\\I=\pi\ Ln2-4*\frac{\pi}{2}Ln2\\\\I=-\ \pi \ Ln2$