The value of $\int_{0}^{4042} \frac{\sqrt{x} \, dx}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2}{1 + \tan x + \sqrt{1 + \tan^2 x}} \, dx$ is

$\int_0^\pi x \sin x \, dx = $

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

The value of $\int_{-1}^{3} \left( \tan^{-1} \left( \frac{x}{x^2+1} \right) + \tan^{-1} \left( \frac{x^2+1}{x} \right) \right) dx$ is

For $n \in N$,if $I_n = \int \frac{\sin nx}{\sin x} dx = \frac{2}{n-1} \sin(n-1)x + I_{n-2}$ and $\int_0^\pi \frac{\sin nx}{\sin x} dx = \frac{k\pi}{2}$,then $k =$