$\int_0^\pi \frac{x \, dx}{1 + \sin x}$ is equal to

Read the following mathematical statements carefully:
$I.$ $A$ differentiable function $f$ with maximum at $x = c$ $\implies f''(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$,$\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx}$.
$IV.$ If $f(x)$ has a maxima at $x = c$,then $f$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \to 0$ for $h > 0$. Now indicate the correct alternative.

$\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) d x=$

Let $f : R \to R$ be a function such that $f(2 - x) = f(2 + x)$ and $f(4 - x) = f(4 + x)$,for all $x \in R$. If $\int_{0}^{2} f(x) dx = 5$,then the value of $\int_{10}^{50} f(x) dx$ is:

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx =$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) dx =$