$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to

The value of $\int_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^5}3x} \,dx$ is

$\int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx =$

Let $g_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}, i=1, 2$,and $f: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}$ be functions such that $g_1(x)=1, g_2(x)=|4x-\pi|$ and $f(x)=\sin^2 x$,for all $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$.
Define $S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, i=1, 2$.
$(1)$ The value of $\frac{16S_1}{\pi}$ is.
$(2)$ The value of $\frac{48S_2}{\pi^2}$ is.

Let $f$ be a non-constant continuous function for all $x \geq 0$. Let $f$ satisfy the relation $f(x) f(a-x)=1$ for some $a \in R^{+}$. Then, $I=\int_{0}^{a} \frac{d x}{1+f(x)}$ is equal to

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x=$