int_(0)^( pi/4)log(1+tan x)dx=(pi)/(8)log2

Prove that int_(0)^(pi/4)log(1+tanx)dx=(pi)/(8) log2.

Prove that int_(0)^((pi)/(4))log(1+tanx)dx=(pi)/(8)log2

8. int_0^(pi/4) log(1+tanx)dx

Using integral int_(0)^(-(pi)/(2))ln(sin x)dx=-int_(0)^( pi)ln(sec x)dx=-(pi)/(2)ln2 and int_(0)^((pi)/(2))ln(tan x)dx=0 and int_(0)^((pi)/(4))ln(1+tan x)dx=(pi)/(8)

Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2

int_(0)^(pi//2) log (tan x ) dx=

int_(0)^((pi)/(2))log(tan x)*dx

int_0^(pi//2) log(tan x)dx =