$\int_0^{\pi / 2} \sin^8 x \, dx =$

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

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ 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 $\mathbb{R}$

Let $f:(0, +\infty) \to \mathbb{R}$ and $F(x) = \int_0^{x^2} f(t) dt$. If $F(x) = x^2(1 + x)$,then $f(4)$ equals

If $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$ where $f$ is a continuous function,then the value of $f(4)$ is: