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

Explanation

माना `I=int_(0)^(pi)log(1+cosx)dx" …(1)"`
`rArr" "I=int_(0)^(pi)log{1+cos(pi-x)}dx`
`" "[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`=int_(0)^(pi)log(1-cosx)dx`
`=int_(0)^(pi)log{2sin^(2)((x)/(2))}dx`
`=int_(0)^(pi){log2+2log(sin.(x)/(2))}dx`
`=int_(0)^(pi)log2dx+2int_(0)^(pi)log(sin.(x)/(2))dx`
दूसरे समाकलन में `(x)/(2)=t` रखने पर ,
`rArr" "dx=2dt" और "x=0" "rArr" "t=0`
तथा `" "x=pi" "rArr" "t=(pi)/(2)`
`therefore" "I=log2[x]_(0)^(pi)+2int_(0)^(pi//2)log (sint)2 dt`
`=log 2(pi-0)+4((-pi)/(2)log2)`
`" "[because int_(0)^(pi//2)log sin x dx =-(pi)/(2)log 2]`
`" "=-pilog2`