$\int_{-5 \pi}^{5 \pi} (1-\cos 2x)^{\frac{5}{2}} dx =$

If $\varphi (x) = \int_{1/x}^{\sqrt{x}} \sin(t^2) \, dt$,then $\varphi'(1) = $

Let $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$. If $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$,then $\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

If $f(x) = \int_{x}^{x^2} (t - 1) dt$,$1 \le x \le 2$,then the global maximum value of $f(x)$ is

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

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