Evaluate: `int_0^pi log(1+cosx)dx`

Answer

`-pilog_(e)2`

Explanation

`I=int_(0)^(pi)log(1+cosx)dx=int_(0)^(pi)log(2"cos"^(2)x/2)dx`
`=int_(0)^(pi)(log2+2"log cos"x/2)dx=pi log 2+2int_(0)^(pi)"log cos"x/2 dx`
`=pi log 2+2xx2int_(0)^(pi//2) log cos t dt, ` where `t=x/2` and `dx=2dt`
`pi log 2=4xx(-(pi)/2 log 2)`
`=-pi log 2`