Hey Aspirant!

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)

Hope this will help you!

All the Best