Let f ( x ) = 0 x g ( t ) log e ( 1 - t 1 + t ) d t , where g is a continuous odd function. If 2 ( f ( x ) + x 2 cos x 1 + e x ) d x, then is equal to [2024]

Question

$Let f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t, where g is a continuous odd function.$

$If \int_{- π / 2}^{π / 2}$ $(f (x) + \frac{x^{2} \cos x}{1 + e^{x}})$ $d x = {(\frac{π}{α})}^{2} - α,$ $then α is equal to_____.$ [2024]

Smart Achievers · Accepted Answer

(2)

Given, $f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t,$ where $g$ is continuous odd function.

$f' (x) = g (x) \log_{e} (\frac{1 - x}{1 + x}) \Rightarrow f' (- x) = g (- x) \log_{e} (\frac{1 + x}{1 - x})$

$= - g (x) [- \log_{e} (\frac{1 - x}{1 + x})] = f (x)$

$∴$ $f' (x)$ is an even function.

$\Rightarrow f (x)$ is an odd function.

Let $I = \int_{- π / 2}^{π / 2} (f (x) + \frac{x^{2} \cos x}{1 + e^{x}}) d x$

$= \int_{- π / 2}^{π / 2} f (x) d x + \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$

$I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$ ...(i) $[∵ \int_{- a}^{a} f (x) d x = 0, as f (x) is an odd function]$

$I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x e^{x}}{1 + e^{x}} d x$ ...(ii)

Adding (i) and (ii), we get

$2 I = \int_{- π / 2}^{π / 2} x^{2} \cos x d x = 2 \int_{0}^{π / 2} x^{2} \cos x d x; I = \int_{0}^{π / 2} x^{2} \cos x d x$

$I = {(x^{2} \sin x)}_{0}^{π / 2} - \int_{0}^{π / 2} 2 x \sin x d x$

$= \frac{π^{2}}{4} - 2 {(- x \cos x)}_{0}^{π / 2} + \int_{0}^{π / 2} \cos x d x = \frac{π^{2}}{4} - 2 {(\sin x)}_{0}^{π / 2}$

$= {(\frac{π}{2})}^{2} - 2 (1 - 0) = {(\frac{π}{2})}^{2} - 2$

$∴ α = 2$

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