The value of $\frac{\left\{\int\limits_{-1 / 2}^{1 / 2} \cos 2 x . \log \left(\frac{1+x}{1-x}\right) d x\right\}}{\left\{\int\limits_0^{1 / 2} \cos 2 x . \log \left(\frac{1+x}{1-x}\right) d x\right\}}$ is
Answer & explanation
Correct answer: option 1
Since $f(x)=\cos 2 x . \log \left(\frac{1+x}{1-x}\right)$ is an odd function.
∴ $\int\limits_{-1 / 2}^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x=0$
Hence, $\frac{\int\limits_{-1 / 2}^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x}{\int\limits_0^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x}=0$