$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x=$ . . . . . . .

The value of the integral $\int_{0}^{\pi}|\sin 2x| dx$ is

Let $f(x)$ be integrable over $(a, b)$,where $b > a > 0$. If $I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \sec^2 \theta \, d\theta$ and $I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \csc^2 \theta \, d\theta$,then the ratio $\frac{I_1}{I_2}$ is:

$\int\limits_a^b [x] \,dx + \int\limits_a^b [-x] \,dx$,where $[.]$ denotes the greatest integer function,is equal to:

$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=$

If $\int \limits_0^\pi \frac{5^{\cos x}(1+\cos x \cos 3x+\cos^2 x+\cos^3 x \cos 3x) dx}{1+5^{\cos x}} = \frac{k \pi}{16}$,then $k$ is equal to $...........$.