Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in \left(0, \frac{\pi}{2}\right)$,if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$,then $f\left(\frac{1}{\sqrt{2}}\right)=$

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

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

Given that $\frac{d}{d x}\left[\int_0^{\phi(x)} f(t) d t\right]=f(\phi(x)) \cdot \phi^{\prime}(x)$. If $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$,then the value of $f(8)$ is

If $f(x) = \int_{x^2}^{x^4} \sin \sqrt{t} \, dt$,then $f'(x)$ equals

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