The value of the integral $\int\limits_{-1 / 2}^{1 / 2} \cos x \log \left(\frac{1+x}{1-x}\right) d x$, is
Answer & explanation
Correct answer: option 1
Let $f(x)=\cos x \log \left(\frac{1+x}{1-x}\right)$. Then,
$f(-x)=\cos (-x) \log \left(\frac{1-x}{1+x}\right)=-\cos x \log \left(\frac{1+x}{1-x}\right)=-f(x)$
$\Rightarrow f(x)$ is an odd function
∴ $\int\limits_{-1/2}^{1/2} \cos x \log (\frac{1+x}{1-x})dx = 0$