If $f(x) = \int_{x^2}^{x^4} \sin \sqrt{t} \, dt$,then $f'(x)$ equals

$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$

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 $\mathbb{R}$

$\int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x=$

Given that $\frac{d}{d x}\left[\int_0^{\phi(x)} f(t) d t\right]=f(\phi(x)) \cdot \phi^{\prime}(x)$. If $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$,then the value of $f(8)$ is

The number of solutions of the equation $\frac{d}{dx} \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}} = 2\sqrt{2}$ in the interval $[0, \pi]$ is: