If $\int_{0}^{\pi/2} \tan^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,then

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.

If $f(x) = \frac{e^x}{1 + e^x}$,$I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$,and $I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$,then the value of $\frac{I_2}{I_1}$ is

If $(a, b)$ is the orthocentre of the triangle whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$,and $I_1 = \int_{a}^{b} x \sin(4x - x^2) dx$,$I_2 = \int_{a}^{b} \sin(4x - x^2) dx$,then $36 \frac{I_1}{I_2}$ is equal to:

$\int_0^\pi \sin^2 x \cos^3 x \, dx = $ . . . . . . .

$\int_0^a f(x) \, dx = $