Evaluate ∫limits_-1/2^1/2 cos x × log ≤ft( (1+x)/(1-x) ) dx

The correct answer is Option (3) → 0

Let $I = \int\limits_{-1/2}^{1/2} \cos x \log \left( \frac{1+x}{1-x} \right) \, dx \quad \dots(i)$

Using property $\int\limits_{a}^{b} f(x) \, dx = \int\limits_{a}^{b} f(a+b-x) \, dx$

$I = \int\limits_{-1/2}^{1/2} \cos \left( -\frac{1}{2} + \frac{1}{2} - x \right) \times \log \left( \frac{1 + (-\frac{1}{2} + \frac{1}{2} - x)}{1 - (-\frac{1}{2} + \frac{1}{2} - x)} \right) \, dx$

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

$I = \int\limits_{-1/2}^{1/2} \cos x \log \left( \frac{1-x}{1+x} \right) \, dx \quad \dots(ii)$

On adding eq. (i) and eq. (ii)

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

Using $\log m + \log n = \log(mn)$:

$2I = \int\limits_{-1/2}^{1/2} \cos x \log \left( \frac{1+x}{1-x} \times \frac{1-x}{1+x} \right) \, dx$

$2I = \int\limits_{-1/2}^{1/2} \cos x \log 1 \, dx$

$2I = \int\limits_{-1/2}^{1/2} \cos x (0) \, dx$

$2I = 0$

$I = 0$