For x ∈ ℝ , let ^ - 1 (x) ∈ (- (π)/(2), (π)/(2)) . Then the minimum...

Given,

$f\left(x\right) = {∫}_{0}^{x\tan^{- 1}x}\dfrac{{e}^{\left(t - \cos t\right)}}{1 + {t}^{2023}}dt$

Now differentiating both side we get,

${f}^{'}\left(x\right) = \dfrac{{e}^{\left[x\tan^{- 1}x - \cos \left(x\tan^{- 1}x\right)\right]}}{1 + {\left(x\tan^{- 1}x\right)}^{2023}} \times \left(\dfrac{x}{1 + {x}^{2}} + \tan^{- 1}x\right)$

$⇒ {f}^{'}\left(x\right) = g\left(x\right) \cdot h\left(x\right)$

$\text{where}g\left(x\right) = \dfrac{{e}^{\left[x\tan^{- 1} x - \cos \left(x\tan^{- 1} x\right)\right]}}{1 + {\left(x\tan^{- 1} x\right)}^{2023}} > 0\forall x$

And h\left(x\right) = \dfrac{x}{1 + {x}^{2}} + \tan^{- 1}x\text{which is}\left\{$\begin{array}{ll} 0x > 0\end{array}$\right)</p><p>∴ f\left(x\right)$has minimum at$x = 0$</p><p>And$f{\left(x\right)}_\min = f\left(0\right) = 0