`int_0^pilog(1+cosx)dx` .

Explanation

`"Let I "= int_(0)^(pi//2) log (1+cos x) dx " "......(1)`
`rArr I =int_(0)^(pi) log{1+cos (pi-x) }dx`
`[ :' int_(0)^(a) f(x) dx= int_(0)^(a) f(a-x) dx]`
`=int_(0)^(pi)log (1-cos x)dx`
`=int_(0)^(pi) log {2sin^(2)((x)/(2))}dx`
` =int_(0)^(pi) {log 2+2 log (sin .(x)/(2))}dx`
`=int_(0)^(pi) log 2 dx+ 2 int_(0)^(pi) log (sin .(x)/(2)) dx`
`"Put " .(x)/(2) =t " in second integral "`
`rArr " "dx=2dt`
`" and " x=0 rArr t=0`
`" and " x=pi rArr t=(pi)/(2)`
`:. I =log 2[x]_(0)^(pi) +2int_(0)^(pi//2) log (sin t) 2dt`
`=log 2(pi-0)+ 4((-pi)/(2)log 2)`
`[ :' int_(0)^(pi//2) log sin x dx =-(pi)/(2)log 2]`
`=-pi log 2`