The derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,$(x > 0)$ 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

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

$\int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x=$

$\int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx =$

Let $S(x) = \int_{x^2}^{x^3} \ln t \, dt$ for $x > 0$ and $H(x) = \frac{S'(x)}{x}$. Then $H(x)$ is :