If $f(x) = \int_{0}^{x} t \sin t \,dt,$ then $f^{\prime}(x)$ is

$\int_0^\pi (\sin^3 x + \cos^2 x)^2 dx = $

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $f(x)=\int_{\frac{1}{x}}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$. Then
$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$,for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $\mathbb{R}$

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 $f, g:(0, \infty) \rightarrow \mathbb{R}$ be two functions defined by $f(x)=\int_{-x}^x(|t|-t^2) e^{-t^2} dt$ and $g(x)=\int_0^{x^2} t^{1/2} e^{-t} dt$. Then the value of $(f(\sqrt{\log_{e} 9}) + g(\sqrt{\log_{e} 9}))$ is

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