$\int_0^{\pi /2} \sin^4 x \cos^6 x \, dx$ equals

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

If $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt} $,then $f'(1/2)$ is

If $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0,$ then $\frac{dy}{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 $\mathbb{R}$