Evaluate `int_(0)^(pi//4)In(1+tanx)dx`

Explanation

Let `I=int_(0)^(pi//4)In(1+tanx)dx`…………..1
`=int_(0)^(pi//4)In(1+tan(pi//4-x))dx`
`[ :'int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`=pi_(0)^(pi//4)In[1+(1-tanx)/(1+tanx)]dx`
`=int_(0)^(pi//4)In(2/(1+tanx))dx`
`I=int_(0)^(pi//4)[(In2-In(1+tanx)]dx`..............2
Adding equations 1 and 2 we get
`2I=int_(0)^(pi//4)In 2 dx`
`=In 2[x]_(0)^(pi//4)=In 2[(pi)/2]`
or `I=(pi)/8 In 2`