$\lim \limits_{x \rightarrow 1} \left( \frac{\int \limits_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)} \right)$ is equal to

For $x \in R$,let $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $f: R \rightarrow R$ defined by $f(x) = \int_0^{x \tan^{-1} x} \frac{e^{(t-\cos x)}}{1+t^{2023}} dt$ is

$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$

Let $f : R \rightarrow R$ be a continuous odd function,which vanishes exactly at one point and $f(1) = \frac{1}{2}$. Suppose that $F(x) = \int_{-1}^x f(t) dt$ for all $x \in [-1, 2]$ and $G(x) = \int_{-1}^x t|f(f(t))| dt$ for all $x \in [-1, 2]$. If $\lim_{x \rightarrow 1} \frac{F(x)}{G(x)} = \frac{1}{14}$,then the value of $f\left(\frac{1}{2}\right)$ is

Let $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,then $\int_0^{\pi /2} f(x) dx = $

Let for some function $y=f(x)$,$\int_0^x t f(t) d t=x^2 f(x)$,$x > 0$ and $f(2)=3$. Then $f(6)$ is equal to :