`int_0^[pi/4]log([sinx+cosx]/cosx)dx`

Explanation

`I = int_0^(pi/4) log((sinx+cosx)/cosx) dx`

`=>I = int_0^(pi/4) log(1+tanx) dx`

We know, ` int_a^(b) f(x)dx = int_a^(b) f(a+b-x) dx`

`:. I = int_0^(pi/4) log(1+tan(pi/4-x) )dx`

`=> I = int_0^(pi/4) log(1+(1-tanx)/(1+tanx)) dx`

`=> I = int_0^(pi/4) log(2/(1+tanx)) dx`

`=> I = int_0^(pi/4) log2 dx - int_0^(pi/4)(1+tanx) dx`

`=>I = pi/4log2 - I`

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