If $I_n = \int_0^a \frac{x^n}{\sqrt{a^2-x^2}} dx$,then $\frac{I_8}{I_4} =$

The value of $\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to:

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

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:

$\lim _{x \rightarrow 0} \frac{48}{x^4} \int _{0}^{x} \frac{t^3}{t^6+1} dt$ is equal to $.......$.