If the value of the integral ∫ _ - (π)/(2) ^ (π)/(2) ( x ^ 2 x 1 +...

Given,

${∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{2}}\left(\dfrac{{x}^{2}\cos x}{1 + {\pi}^{x}} + \dfrac{1 + \sin^{2}x}{1 + {e}^{\sin^{2023}x}}\right)dx = \dfrac{\pi}{4}\left(\pi + \alpha\right) - 4$

Now let, ${I}_{1} = {∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{2}}\left(\dfrac{{x}^{2}\cos x}{1 + {\pi}^{x}}\right)dx . . . . . . . \left(i\right)$

Now using the property ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{b}f\left(a + b - x\right) d x$ we get,

$⇒ {I}_{1} = {∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{2}}\left(\dfrac{{x}^{2}\cos x}{1 + {\pi}^{- x}}\right)dx . . . . . . . .$\left(ii\right)$</p><p>Now adding above equations we get,</p><p>$⇒ 2{I}_{1} = 2{∫}_{0}^{\dfrac{\pi}{2}}\left({x}^{2}\cos x\right)dx$\left\{\text{as}{∫}_{- a}^{a}f\left(x\right) d x = 2{∫}_{0}^{a}f\left(x\right) d x, \text{if}f\left(x\right)\text{is even}\right\}$

$⇒ {I}_{1} = {\left[{x}^{2}\left(\sin x\right)\right]}_{0}^{\dfrac{\pi}{2}} - {∫}_{0}^{\dfrac{\pi}{2}}2x \cdot sinxdx$

$⇒ {I}_{1} = \dfrac{{\pi}^{2}}{4} - \left[{\left[- 2xcosx\right]}_{0}^{\dfrac{\pi}{2}} + {∫}_{0}^{\dfrac{\pi}{2}}2cosxdx\right]$

$⇒ {I}_{1} = \dfrac{{\pi}^{2}}{4} - 2$

Similarly solving, ${I}_{2} = {∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{2}}\left(\dfrac{1 + \sin^{2}x}{1 + {e}^{\sin^{2023}x}}\right)dx$

$⇒ {I}_{2} = {∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \sin^{2}x\right)dx$

$⇒ {I}_{2} = {∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \dfrac{1 - \cos 2x}{2}\right)dx$

$⇒ {I}_{2} = \dfrac{1}{2}{∫}_{0}^{\dfrac{\pi}{2}}\left(3 - \cos 2x\right)dx$

$⇒ {I}_{2} = \dfrac{1}{2}{\left[3x - \dfrac{\sin 2x}{2}\right]}_{0}^{\dfrac{\pi}{2}}$

$⇒ {I}_{2} = \dfrac{3\pi}{4}$

$⇒ {I}_{1} + {I}_{2} = \dfrac{{\pi}^{2}}{4} - 2 + \dfrac{3\pi}{4}$

$⇒ \dfrac{\pi}{4}\left(\pi + 3\right) - 2 = \dfrac{\pi}{4}\left(\pi + \alpha\right) - 2$

$⇒ \alpha = 3$