Question

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

Answer 1

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