Find $\min \int_0^{x\arctan x}\frac{e^{t-\cos t}}{1+t^{2023}}\mathrm dt$

Question

Find the minimum value of : $$f(x) = \int_0^{x\arctan x}\frac{e^{t-\cos t}}{1+t^{2023}}\mathrm dt$$

Using FTC, we have ;

$$f'(x) = \frac{e^{x\arctan x - \cos (x\arctan x)}}{1+x\arctan x}.(\arctan x + \frac{x}{x^2 + 1})$$

For $f$ to be minimum, $f' = 0\implies\arctan x = -\frac{x}{x^2 + 1}$

However, I can't proceed further and don't find it much useful.

This question appeared today in my test (JEE) for which I had an average of 3.5 minutes to solve it. But in such short time, I am failing to find a short and elegant method.

Answer 1

Find the minimum value of :$$f(x)=\int_0^{x\arctan x}\frac{e^{t-\cos t}}{1+t^{2023}}\,\mathrm dt$$

Since $\;x\arctan x>0\;$ for any $\;x\neq0\;$ and $\;\dfrac{e^{t-\cos t}}{1+t^{2023}}>0\;$ for any $\;t\geqslant0\,,\;$ it follows that

$f(x)>0\;$ for any $\;x\neq0\,,\;$ hence

the minimum value of $\,f(x)\,$ is

$f(0)=0\,$.