The value of `I=int_0^1 ln(cosx)dx` has the same value as that ofthe definite integral below

int_0^pi log(1+cosx)dx

Find I=int_0^pi ln(1+cosx)dx

Find I=int_0^pi ln(1+cosx)dx

Evaluate: int_0^pi log(1+cosx)dx

Evaluate: int_0^pi log(1+cosx)dx

Evaluate: int_0^pi log(1+cosx)dx

The value of I=int_(0)^(pi//2) (1)/(1+cosx)dx is

The value of I=int_(0)^(pi//2) (1)/(1+cosx)dx is

Given int_(0)^(pi//2)(dx)/(1+sinx+cosx)=A . Then the value of the definite integral int_(0)^(pi//2)(sinx)/(1+sinx+cosx)dx is equal to

Given int_(0)^(pi//2)(dx)/(1+sinx+cosx)=A . Then the value of the definite integral int_(0)^(pi//2)(sinx)/(1+sinx+cosx)dx is equal to