Question

The value of the integral
${\int }_{-1/2}^{1/2}\cos x.\log \left(\frac{1+x}{1-x}\right)dx$ is

• A. 0
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A 0

$={\int }_{-1/2}^{1/2}cosxlog\left(\frac{1+x}{1-x}\right)dx$
$I={\int }_{-1/2}^{-1/2}cos(-x)log\left(\frac{1-x}{1+x}\right)dx$
$={\int }_{-1/2}^{-1/2}cosxlog\left(\frac{1-x}{1+x}\right)dx$
$2I={\int }_{-1/2}^{1/2}0dx$
$I=0$