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

If $f(x) = \int_{x}^{x^2} (t - 1) dt$,$1 \le x \le 2$,then the global maximum value of $f(x)$ is

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x}{3}\right) d\lambda$ for $x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$,then $\alpha$ is equal to $.........$

$\int\limits_0^{\frac{\pi }{4}} (\cos 2x)^{3/2} \cos x \,dx =$

If $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ for all $x \in R - \{0\}$,then the value of $f(\frac{\pi}{6})$ 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 $\mathbb{R}$