Question

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

Answer 1

$f(x)$ has minimum at $x=0$
And $f(x)_{min}=f(0)$
$f(x)_{min}=0$

So, the answer is $0$.