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 $f\left(\frac{1}{\sqrt{2}}\right)=$

Let $f$ be a continuous function satisfying $\int \limits_0^{t^2} (f(x) + x^2) dx = \frac{4}{3} t^3, \forall t > 0$. Then $f \left(\frac{\pi^2}{4}\right)$ is equal to:

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ is

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

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

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