Explanation

(a)
Let $I = \int_{- 1/2}^{1/2}{\cos{x\log{\left( \frac{1 - x}{1 + x} \right)dx}}}$ …(i)
On replacing $x\ by - x,$ we get
$I = \int_{- 1/2}^{1/2}{\cos{( - x)\log{\left( \frac{1 + x}{1 - x} \right)dx}}}$
$\Rightarrow \ I = - \int_{- 1/2}^{1/2}{\cos{x\log{\left( \frac{1 - x}{1 + x} \right)dx}}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{- 1/2}^{1/2}{\cos{x\log{\left( \frac{1 - x}{1 + x} \right)dx - \int_{- 1/2}^{1/2}{\cos{x\log\left( \frac{1 - x}{1 + x} \right)}}}}}dx$
$\Rightarrow \ 2I = 0$
$\Rightarrow \ I = 0$