- 1 / 2 1 / 2 ( cos x ) [ log ( 1 - x 1 + x ) ] d x =

Question

$\int_{- 1 / 2}^{1 / 2} (\cos x) [\log (\frac{1 - x}{1 + x})] d x =$

Smart Achievers · Accepted Answer

(1)

Ans. 0

Explanation :

$I = \int_{- 1 / 2}^{1 / 2} (\cos x) [\log (\frac{1 - x}{1 + x})] d x$ $\dots (i)$

$\Rightarrow$ $I = \int_{- 1 / 2}^{1 / 2} \cos (- x) [\log (\frac{1 + x}{1 - x})] d x$ $[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$

$\Rightarrow$ $I = - \int_{- 1 / 2}^{1 / 2} \cos x [\log (\frac{1 - x}{1 + x})] d x$ $\dots (i i)$

$Adding (i) and (ii), we get$

$2 I = \int_{- 1 / 2}^{1 / 2} \cos x [\log (\frac{1 - x}{1 + x})] d x - \int_{- 1 / 2}^{1 / 2} \cos x [\log (\frac{1 - x}{1 + x})] d x$

$or$ $2 I = 0$ $or$ $I = 0$ $[∵ \log (1) = 0]$

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