सिद्ध कीजिए कि
`int_(0)^(pi//8)log(1+tan2x)dx=(pi)/(16)log_(e)2`

Explanation

माना `I=int_(0)^(pi//8)log(1+tan 2x)dx" …(1)"`
`rArr" "I=int_(0)^(pi//8)log{1+tan2((pi)/(8)-x)}dx`
`=int_(0)^(pi//8)log{1+tan((pi)/(4)-2x)}dx`
`=int_(0)^(pi//8)log{1+((1-tan2x)/(1+tan2x))}dx`
`=int_(0)^(pi//8)log((1)/(1+tan2x))dx`
`=int_(0)^(pi//8)log2dx-int_(0)^(pi//8)log(1+tan2x)dx`
`rArr" "I=log2[x]_(0)^(pi//8)-I`
`rArr" "2I=(pi)/(8)log2" "rArr" "I=(pi)/(16)log2`