$\int_0^{\pi /2} \sin^2 x \cos^3 x \, dx = $

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ is equal to

$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)+\sin ^5\left(\frac{3 \pi}{6 n}\right)+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\} = $

$\int_{-2}^2 x^4(4-x^2)^{\frac{7}{2}} dx=$

The angle of intersection between the curves $y = \int\limits_{x^2}^{x^3} \sqrt{5 - t^2} \, dt$ and the $x$-axis is (where $x \neq 0$):

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 $\mathbb{R}$