$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

$\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) dt }{x^{3}}$ is equal to

The value of the integral $\int_0^{\pi / 2} \sin^5 x \, dx$ is

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 $\mathbb{R}$

$\int_{-2}^2 (4-x^2)^{\frac{5}{2}} dx = $ (in $\pi$)

If $f(x) = \int_0^x {t\sin t\,dt} $,then $f'(x) = $