$\int\limits_{0}^{\pi} |\sin 2x|\, \text{d}x =$

If $f(x)$ is an even function, then $\int\limits_{-2}^{2} (|x| + f(x)\sin x)\, \text{d}x$ is

$\int_0^{\log 10}\left[\dfrac{e^x \sqrt{e^x - 1}}{e^x + 8}\right]dx = $

The value of the integral $\int_{-\pi/2}^{\pi/2}\left(\dfrac{x^2 \cos x}{1+e^x}\right)dx$ is equal to $\left(\dfrac{\pi^2}{A}\right) - B$. Then $\left(\dfrac{A}{B}\right) = $