For $x \in \mathbb{R}$, let $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \int_{0}^{x \tan^{-1} x} \frac{e^{(t-\cos t)}}{1+t^{2023}}\,dt$ is

For x ∈ R, let tan⁻¹(x) ∈ (-π/2, π/2). Then the minimum value of the function f : R → R defined by f(x) = ∫₀^(x tan⁻¹ x) (e^((t-cos t)))/(1+t²⁰²³) dt is

$f(x) = \int_{0}^{x\tan^{-1}x}\frac{e^{t-\cos t}}{1+t^{2023}}\,dt$

$\tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$

$\text{Find the minimum value of } f$

$1.\ \text{sign of upper limit } u(x)=x\tan^{-1}x$

$2.\ \text{sign of the integrand}$

$3.\ \text{locate the minimum}$

$u(x) = x\tan^{-1}x \ge 0$

$u(x)=0 \iff x=0$

$g(t) = \frac{e^{t-\cos t}}{1+t^{2023}} > 0\ \text{for } t\ge 0$

$f(x) = \int_0^{u(x)} g(t)\,dt \ge 0$

$f(x)=0 \iff u(x)=0 \iff x=0$

$f'(x) = g(u(x))\,u'(x),\quad g>0$

$\min u = u(0) = 0 \Rightarrow \min f = f(0)$

$f(0) = \int_0^0 g(t)\,dt = 0$

$\boxed{\text{Minimum value} = 0}$