The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \frac{\pi-x}{\pi+x}\right) \cos x \, dx$ 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.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ is

$\int_{-\pi / 15}^{\pi / 15} \frac{\cos 5 x}{1+e^{5 x}} d x=$

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

The value of $\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x$ is equal to